Cww97 Code Library 2017.07.05
East China Normal University
Chen WeiWen - Software Engineering
Cao ZhiJie - Computer Science
Zhu XuLiang - Mathematics
常用STL
map的Upperbound
map<int,int>::iterator se = mp.upper_bound(mid); 返回迭代器
优先队列
priority_queue<int>Q;//采用默认优先级构造队列 priority_queue<int,vector<int>,cmp1>que1;//最小值优先 priority_queue<int,vector<int>,cmp2>que2;//最大值优先 Q.push(x); int x = Q.top(); Q.pop();
multiset
begin() 返回指向第一个元素的迭代器 clear() 清除所有元素 count() 返回某个值元素的个数 empty() 如果集合为空，返回 true end() 返回指向最后一个元素的迭代器 erase() 删除集合中的元素 ( 参数是一个元素值，或者迭代器) find() 返回一个指向被查找到元素的迭代器 insert() 在集合中插入元素 size() 集合中元素的数目 lower_bound() 返回指向大于（或等于）某值的第一个元素的迭代器 upper_bound() 返回大于某个值元素的迭代器 equal_range() 返回集合中与给定值相等的上下限的两个迭代器 multiset <point> po; multiset <point>::iterator L, R, it;
离散化lower_lound
sort(a + 1, a + A+1);
A = unique(a + 1, a + A+1) - (a + 1);
for (int i = 1; i <= n; i++){ // segtree
int L = lower_bound(a+1, a+A+1, l[i]) - a;
int R = lower_bound(a+1, a+A+1, r[i]) - a;
T.update(L, R, i, 1, T.M+1,1);
}
------------use in ChairTree-------------------
for (int i = 1; i <= n; i++) {
scanf("%d", &arr[i]);
Rank[i] = arr[i];
}
sort(Rank + 1, Rank + n+1);//Rank存储原值
int m = unique(Rank + 1, Rank + n +1) - (Rank + 1);//这个m很重要，WA一天系列
for (int i = 1; i <= n; i++) {//离散化后的数组，仅仅用来更新
arr[i] = lower_bound(Rank + 1, Rank + m+1, arr[i]) - Rank;
}
==================CZJ排序去重===================
sort(vecs.begin(), vecs.end());
vecs.resize(unique(vecs.begin(), vecs.end()) - vecs.begin());
bitset
构造函数
bitset<n> b;
b有n位，每位都为0.参数n可以为一个表达式.
如bitset<5> b0;则"b0"为"00000";
bitset<n> b(unsigned long u);
b有n位,并用u赋值;如果u超过n位,则顶端被截除
如:bitset<5>b0(5);则"b0"为"00101";
bitset<n> b(string s);
b是string对象s中含有的位串的副本
string bitval ( "10011" );
bitset<5> b0 ( bitval4 );
则"b0"为"10011";
bitset<n> b(s, pos);
b是s中从位置pos开始位的副本,前面的多余位自动填充0;
string bitval ("01011010");
bitset<10> b0 ( bitval5, 3 );
则"b0" 为 "0000011010";
bitset<n> b(s, pos, num);
b是s中从位置pos开始的num个位的副本,如果num<n,则前面的空位自动填充0;
string bitval ("11110011011");
bitset<6> b0 ( bitval5, 3, 6 );
则"b0" 为 "100110";
bool any() 是否存在置为1的二进制位？和none()相反
bool none() 是否不存在置为1的二进制位,即全部为0？和any()相反.
size_t count() 二进制位为1的个数.
size_t size() 二进制位的个数
flip() 把所有二进制位逐位取反
flip(size_t pos)把在pos处的二进制位取反
bool operator[]( size_type _Pos )获取在pos处的二进制位
set() 把所有二进制位都置为1
set(pos) 把在pos处的二进制位置为1
reset() 把所有二进制位都置为0
reset(pos) 把在pos处的二进制位置为0
test(size_t pos)在pos处的二进制位是否为1？
unsigned long to_ulong( )用同样的二进制位返回一个unsigned long值
string to_string ()返回对应的字符串.
线性代数
矩阵快速幂
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long LL;
const LL MOD = 1000000007LL;
LL n,m;
struct matrix{
static const int MATRIX_N = 11;
LL a[MATRIX_N][MATRIX_N];
int row, col;
matrix():row(MATRIX_N),col(MATRIX_N){memset(a,0,sizeof(a));}
matrix(int x, int y):row(x),col(y){memset(a,0,sizeof(a));}
LL* operator [] (int x){return a[x];}
matrix operator * (matrix x){
matrix tmp(col, x.row);
for(int i = 0; i < row; i++)
for(int j = 0; j < col; j++) if(a[i][j])//稀疏矩阵优化
for(int k = 0; k < x.col; k++) if (x[j][k]){
tmp[i][k] += a[i][j] * x[j][k];
//mult(a[i][j], x[j][k], MOD);
tmp[i][k] %= MOD;
}
return tmp;
}
void operator *= (matrix x){*this = *this * x;}
matrix operator ^ (LL x){
matrix ret(row, col);
for (int i = 0; i < col; i++) ret[i][i] = 1;
matrix tmp = *this;
for (; x; x >>= 1, tmp *= tmp){if (x&1) ret *= tmp;}
return ret;
}
void print(){
for (int i = 0; i < row; i++){
for (int j = 0; j < col; j++) printf("%lld ",a[i][j]);
puts("");
}
}
};
int main() {
LL n;
matrix B(3, 1);
while(scanf("%lld%lld%lld%lld", &B[0][0], &B[1][0], &B[2][0], &n) == 4) {
matrix A(3, 3);
A[0][0] = 0; A[0][1] = 1; A[0][2] = 0;
A[1][0] = 0; A[1][1] = 0; A[1][2] = 1;
A[2][0] = 0; A[2][1] = 1; A[2][2] = 1;
A = A ^ n;
matrix C(3, 1);
C = A * B;
printf("%lld\n", (C[0][0] + C[1][0] + C[2][0]) % MOD);
}
return 0;
}
就是快速幂(czj)
LL fast_multi(LL m, LL n, LL mod){//快速乘法
LL ans = 0;//注意初始化是0，不是1
for (;n; n >>= 1){
if (n & 1) ans += m;
m = (m + m) % mod;//和快速幂一样，只不过这里是加
m %= mod;//取模，不要超出范围
ans %= mod;
}
return ans;
}
LL fast_pow(LL a, LL n, LL mod){//快速幂
LL ans = 1;
for (;n;n >>= 1){
if (n & 1) ans = fast_multi(ans, a, mod);//不能直接乘
a = fast_multi(a, a, mod);
ans %= mod;
a %= mod;
}
return ans;
}
线性递推函数杜教模板
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// head
int _;
ll n;
namespace linear_seq {
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<ll> Md;
void mul(ll *a,ll *b,ll k) {
rep(i,0,k+k) _c[i]=0;
rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--) if (_c[i])
rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
rep(i,0,k) a[i]=_c[i];
}
int solve(ll n,VI a,VI b) {
ll ans=0,pnt=0;
ll k=SZ(a);
assert(SZ(a)==SZ(b));
rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
Md.clear();
rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
rep(i,0,k) res[i]=base[i]=0;
res[0]=1;
while ((1ll<<pnt)<=n) pnt++;
for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
}
}
rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
rep(n,0,SZ(s)) {
ll d=0;
rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m) C.pb(0);
rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
int gao(VI a,ll n) {
VI c=BM(a);
c.erase(c.begin());
rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
int main() {
for (scanf("%d",&_);_;_--) {
scanf("%lld",&n);
printf("%d\n",linear_seq::gao(VI{31, 197, 1255, 7997, 50959, 324725, 2069239, 13185773, 84023455},n-2));
}
}
高斯消元（naive）
判断是否有解
求解见红书
int Gauss(matrix a, int m, int n){
int x_cnt = 0;
int col, k; //col为列号,k为行号
for (k=0,col=0;k<m&&col<n; ++k, ++col){
int r = k; //r为第col列的一个1
for (int i=k;i<m;++i) if (a[i][col])r=i;
if (!a[r][col]){ k--; continue;}
if (r!=k)for (int i=col;i<=n;++i)
swap( a[r][i], a[k][i]);
for (int i=k+1;i<m; ++i)if (a[i][col])//消元
for (int j=col;j<=n;++j) a[i][j] ^= a[k][j];
}
for (int i=k;i<m;++i) if (a[i][n])return -1;
if (k<=n)return n-k; //返回自由元个数
}
NTT（对任意数取模）
// 多项式乘法 系数对MOD=1000000007取模， 常数巨大，慎用
// 只要选的K个素数乘积大于MOD*MOD*N,理论上MOD可以任取。
#define MOD 1000000007
#define K 3
const int m[K] = {1004535809, 998244353, 104857601};
#define G 3
int qpow(int x, int k, int P) {
int ret = 1;
while(k) {
if(k & 1) ret = 1LL * ret * x % P;
k >>= 1;
x = 1LL * x * x % P;
}
return ret;
}
struct _NTT {
int wn[25], P;
void init(int _P) {
P = _P;
for(int i = 1; i <= 21; ++i) {
int t = 1 << i;
wn[i] = qpow(G, (P - 1) / t, P);
}
}
void change(int *y, int len) {
for(int i = 1, j = len / 2; i < len - 1; ++i) {
if(i < j) swap(y[i], y[j]);
int k = len / 2;
while(j >= k) {
j -= k;
k /= 2;
}
j += k;
}
}
void NTT(int *y, int len, int on) {
change(y, len);
int id = 0;
for(int h = 2; h <= len; h <<= 1) {
++id;
for(int j = 0; j < len; j += h) {
int w = 1;
for(int k = j; k < j + h / 2; ++k) {
int u = y[k];
int t = 1LL * y[k+h/2] * w % P;
y[k] = u + t;
if(y[k] >= P) y[k] -= P;
y[k+h/2] = u - t + P;
if(y[k+h/2] >= P) y[k+h/2] -= P;
w = 1LL * w * wn[id] % P;
}
}
}
if(on == -1) {
for(int i = 1; i < len / 2; ++i) swap(y[i], y[len-i]);
int inv = qpow(len, P - 2, P);
for(int i = 0; i < len; ++i)
y[i] = 1LL * y[i] * inv % P;
}
}
void mul(int A[], int B[], int len) {
NTT(A, len, 1);
NTT(B, len, 1);
for(int i = 0; i < len; ++i) A[i] = 1LL * A[i] * B[i] % P;
NTT(A, len, -1);
}
}ntt[K];
int tmp[N][K], t1[N], t2[N];
int r[K][K];
int CRT(int a[]) {
int x[K];
for(int i = 0; i < K; ++i) {
x[i] = a[i];
for(int j = 0; j < i; ++j) {
int t = (x[i] - x[j]) % m[i];
if(t < 0) t += m[i];
x[i] = 1LL * t * r[j][i] % m[i];
}
}
int mul = 1, ret = x[0] % MOD;
for(int i = 1; i < K; ++i) {
mul = 1LL * mul * m[i-1] % MOD;
ret += 1LL * x[i] * mul % MOD;
if(ret >= MOD) ret -= MOD;
}
return ret;
}
void mul(int A[], int B[], int len) {
for(int id = 0; id < K; ++id) {
for(int i = 0; i < len; ++i) {
t1[i] = A[i];
t2[i] = B[i];
}
ntt[id].mul(t1, t2, len);
for(int i = 0; i < len; ++i)
tmp[i][id] = t1[i];
}
for(int i = 0; i < len; ++i) {
A[i] = CRT(tmp[i]);
}
}
void init() {
for(int i = 0; i < K; ++i) {
for(int j = 0; j < i; ++j) {
r[j][i] = qpow(m[j], m[i] - 2, m[i]);
}
}
for(int i = 0; i < K; ++i) {
ntt[i].init(m[i]);
}
}
数论
常用公式和找规律
斯特林公式
n 约等于 sqrt(2*pi*n)*pow(1.0*n/e,n)
带标号连通图计数
1 1 1 4 38 728 26704 1866256 251548592
h(n)=2^(n(n-1)/2)
f(n) = h(n)-sum{C(n-1,k-1)*f(k)*h(n-k)}(k=1...n-1)
不带标号n个节点的有根树计数
1, 1, 2, 4, 9, 20, 48, 115, 286, 719, 1842,
不带标号n个节点的树的计数
1,2,3,6,11,23,47,106,235
OEIS
A(x) = 1 + T(x) - T^2(x)/2 + T(x^2)/2, where T(x) = x + x^2 + 2*x^3 + ... is the g.f. for A000081
错排公式
D[1] = 0; D[2] = 1;
for(int i = 3; i < 25; i++) {
D[i] = (i - 1) * (D[i - 1] + D[i - 2]);
}
卡特兰数
1 2 5 14 42 132 429 1430 4862 16796
binomial(2*n, n)-binomial(2*n, n-1)
Sum_{k=0..n-1} a(k)a(n-1-k)
Stirling数，又称为斯特灵数。
在组合数学，Stirling数可指两类数，都是由18世纪数学家James Stirling提出的。
第一类Stirling数是有正负的，其绝对值是包含n个元素的集合分作k个环排列的方法数目。
第二类Stirling数是把包含n个元素的集合划分为正好k个非空子集的方法的数目。
递推公式
第一类Stirling数是有正负的，其绝对值是包含n个元素的集合分作k个环排列的方法数目。
递推公式为，
S(n,0) = 0, S(1,1) = 1.
S(n 1,k) = S(n,k-1) nS(n,k)。
第二类Stirling数是把包含n个元素的集合划分为正好k个非空子集的方法的数目。
递推公式为：
S(n,k)=0; (n<k||k=0)
S(n,n) = S(n,1) = 1,
S(n,k) = S(n-1,k-1) kS(n-1,k).
第一类斯特林数
有符号Stirling数（无符号Stirling数直接取绝对值）
n=0 1
n=1 0 1
n=2 0 -1 1
n=3 0 2 -3 1
n=4 0 -6 11 -6 1
n=5 0 24 -50 35 -10 1
n=6 0 -120 274 -225 85 -15 1
n=7 0 720 -1764 1624 -735 175 -21 1
第二类
n=0 1
n=1 0 1
n=2 0 1 1
n=3 0 1 3 1
n=4 0 1 7 6 1
n=5 0 1 15 25 10 1
n=6 0 1 31 90 65 15 1
n=7 0 1 63 301 350 140 21 1
n=8 0 1 127 966 1701 1050 266 28 1
n=9 0 1 255 3025 7770 6951 2646 462 36 1
欧拉筛
#include <cstring>
using namespace std;
int prime[1100000],primesize,phi[11000000];
bool isprime[11000000];
void getlist(int listsize){
memset(isprime, 1, sizeof(isprime));
isprime[1] = false;
for(int i=2;i<=listsize;i++){
if(isprime[i]) prime[++primesize]=i;
for(int j = 1; j <= primesize && i*prime[j] <= listsize;j++){
isprime[i*prime[j]] = false;
if(i%prime[j] == 0) break;
}
}
}
莫比乌斯函数模板
void Init(){
memset(vis,0,sizeof(vis));
mu[1] = 1;
cnt = 0;
for(int i=2; i<N; i++){
if(!vis[i]){
prime[cnt++] = i;
mu[i] = -1;
}
for(int j=0; j<cnt&&i*prime[j]<N; j++){
vis[i*prime[j]] = 1;
if(i%prime[j]) mu[i*prime[j]] = -mu[i];
else{
mu[i*prime[j]] = 0;
break;
}
}
}
}
乘法逆元
//扩展欧几里得（扩展gcd）
LL ex_gcd(LL a,LL b,LL &x,LL &y){
if (a == 0 && b == 0) return -1;
if (b == 0){x = 1; y = 0; return a;}
LL d=ex_gcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
LL mod_inverse(LL a,LL n){//乘法逆元
LL x,y;
LL d = ex_gcd(a,n,x,y);
return (x % n + n) % n;
}
//p是质数可以 快速幂p-2
LL quick_mod(LL a, LL b){
LL ans = 1;
a %= MOD;
for(;b; b >>= 1, a = a*a % MOD){
if(b & 1) ans = ans * a % MOD;
}
return ans;
}
//逆元筛：求1-MAXN的所有关于MOD的逆元
inv[0] = 0; inv[1] = 1;
for(i = 2; i < MAXN; i++){
inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
f[i] = (f[i-1] * i) % MOD;
}
详解ACM组合数处理，
O(n2)算法——杨辉三角
O(n)算法——阶乘取模 + 乘法逆元
C(m,n) = n! / m! / (n - m)!
如果p是质数，直接quick_mod(b, p-2) % p 费马小定理求逆元
LL C(LL n, LL m){
if(m > n) return 0;
LL ans = 1;
for(int i = 1; i <= m; i++){
LL a = (n + i - m) % MOD;
LL b = i % MOD;
ans = ans * (a * quick_mod(b, p-2) % MOD) % MOD;
}
return ans;
}
如果n,m很大 达到1e18，但是p很小 <= 1e5 ，我们可以利用这个p
Lucas定理：C(n, m) % p = C(n / p, m / p) * C(n%p, m%p) % p
LL Lucas(LL n, LL m){
if(m == 0) return 1;
return C(n % p, m % p) * Lucas(n / p, m / p) % p;
}
void InitFac(){//阶乘预处理
fac[0] = 1;
for(int i=1; i<=n; i++)
fac[i] = (fac[i-1] * i) % MOD;
}
LL C(LL n,LL m,LL p,LL fac[]){
if(n < m) return 0;
return fac[n] * quick_mod(fac[m] * fac[n-m], p - 2, p) % p;
}
组合数奇偶性结论：
如果（n&m) == m 那么c(m,n)是奇数，否则是偶数
预处理所有数的因数
//预处理所有数的因数表
//SPEED: ECNUOJ 1E6 5000MS O(nlogn)
const int N = 100000 + 5;
vector<int > factor[N];
void init(){
for(int i = 2; i < N; i ++){
for(int j = i; j < N; j += i){
factor[j].push_back(i);
}
}
}
//预处理质因数表
vector<int> x[N];
bool is[N];
void prime() {
memset(is, false, sizeof(is));
for (int i=0; i<N; i++) x[i].clear();
for (int j=2; j<N; j+=2) x[j].push_back(2);
for (int i=3; i<N; i+=2)
if (!is[i]) {
for (int j=i; j<N; j+=i) {
is[j] = true;
x[j].push_back(i);
}
}
}
随机素数测试和大数分解(POJ 1811)
/***************************************************
* Miller_Rabin 算法进行素数测试
* 速度快，可以判断一个 < 2^63 的数是不是素数
****************************************************/
const int S = 8; //随机算法判定次数，一般8~10就够了
// 快速乘法，计算ret = (a*b)%c a,b,c < 2^63
long long mult_mod(long long a,long long b,long long c)
// 快速幂，计算 ret = (a^n)%mod
long long pow_mod(long long a,long long n,long long mod)
// 通过 a^(n-1)=1(mod n)来判断n是不是素数
// n-1 = x*2^t 中间使用二次判断
// 是合数返回true, 不一定是合数返回false
bool check(long long a,long long n,long long x,long long t){
long long ret = pow_mod(a,x,n);
long long last = ret;
for(int i = 1; i <= t; i++){
ret = mult_mod(ret,ret,n);
if(ret == 1 && last != 1 && last != n-1)return true;//合数
last = ret;
}
if(ret != 1)return true;
else return false;
}
//**************************************************
// Miller_Rabin算法
// 是素数返回true,(可能是伪素数)
// 不是素数返回false
//**************************************************
bool Miller_Rabin(long long n){
if (n < 2) return false;
if (n == 2) return true;
if ((n&1) == 0) return false;//偶数
long long x = n - 1, t = 0;
for(; (x&1)==0;){x >>= 1; t++;}
srand(time(NULL));/* *************** */
for(int i = 0; i < S; i++){
long long a = rand()%(n-1) + 1;
if( check(a,n,x,t) ) return false;
}
return true;
}
//**********************************************
// pollard_rho 算法进行质因素分解
//*********************************************
long long factor[100];//质因素分解结果（刚返回时时无序的）
int tol;//质因素的个数，编号0~tol-1
long long gcd(long long a,long long b)
//找出一个因子
long long pollard_rho(long long x,long long c){
long long i = 1, k = 2;
srand(time(NULL));
long long x0 = rand()%(x-1) + 1;
long long y = x0;
while(1){
i ++;
x0 = (mult_mod(x0,x0,x) + c)%x;
long long d = gcd(y - x0,x);
if( d != 1 && d != x)return d;
if(y == x0) return x;
if(i == k){y = x0; k += k;}
}
}
//对 n进行素因子分解，存入factor. k设置为107左右即可
void findfac(long long n,int k){
if(n == 1)return;
if(Miller_Rabin(n)){
factor[tol++] = n;
return;
}
long long p = n;
int c = k;
while( p >= n) p = pollard_rho(p,c--);//值变化，防止死循环k
findfac(p,k);
findfac(n/p,k);
}
//POJ 1811
//给出一个N(2 <= N < 2^54),如果是素数，输出"Prime",否则输出最小的素因子
int main(){
int T; scanf("%d",&T); long long n;
while(T--){
scanf("%I64d",&n);
if(Miller_Rabin(n)) printf("Prime\n");
else{
tol = 0;
findfac(n,107);
long long ans = factor[0];
for(int i = 1; i < tol; i++)
ans = min(ans,factor[i]);
printf("%I64d\n",ans);
}
}
return 0;
}
中国剩余定理
//可以不满足两两互质
int n;
//扩展gcd多了一个变量
void ex_gcd(LL a, LL b, LL &d, LL &x, LL &y){
if (!b) {d = a, x = 1, y = 0;}
else{
ex_gcd(b, a % b, d, y, x);
y -= x * (a / b);
}
}
//注意M的范围
//如果M是int，都要开ll，M是ll，乘法要用快速乘
LL ex_crt(LL *m, LL *r, int n){
LL M = m[1], R = r[1], x, y, d;
for (int i = 2; i <= n; ++i){
ex_gcd(M, m[i], d, x, y);
if ((r[i] - R) % d) return -1;
x = (r[i] - R) / d * x % (m[i] / d);
R += x * M;
M = M / d * m[i];
R %= M;
}
return R > 0 ? R : R + M;
}
离散对数（关于方程x^A=B(mod C)的解)
首先判断是否有解，即 a,p 是否互质。不互质即无解。不妨令 x = im − j, 其中
m = ⌈ √ q⌉ , 这样问题变为求得一组 i j 使得条件满足。此时原式变为 a im−j ≡ b (Mod p), 移
项化简得 (a m ) i ≡ ba j (Mod p)。这个时候我们只需穷举 i,j 使得式子成立即可。先从让 j 从
[0,m] 中穷举，并用 hash 记录下 ba j 对应的 j 值。相同的 ba j 记录较大的 j. 接着让 i 从 [1,m]
中穷举，如果 (a m ) i 在 hash 表中有对应的 j 存在，则对应的 im − j 是一组解。其中第一次出
现的为最小的解。
map<LL, int> Hash;
LL i, j;
LL bsgs(LL a, LL b, LL p){
LL xx, yy;
if (exgcd(a, p, xx, yy) != 1)return −1;
a %= p;
LL m = ceil(sqrt(p));
Hash.clear();
LL tmp, ans = b % p;
for (int i = 0; i <= m; ++i){
Hash[ans] = i;
ans = ans * a % p;
}
tmp = f(a, m, p);
ans = 1;
for (int i = 1; i <= m; ++i){
ans = ans * tmp % p;
if (Hash[ans] != 0) return i * m − Hash[ans];
}
return −1;
}
生成函数 五边形数定理 整数拆分模板
hdu4658
问一个数n能被拆分成多少种方法，且每一种方法里数字重复个数不能超过k（等于k）。
f[n]=∑(-1)^(k-1)(f[n-k(3*k-1)/2]+f[n-k*(3*k+1)/2])
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int N = 100005;
const int MOD = 1000000007;
int dp[N];
void Init() {
dp[0] = 1;
for(int i=1;i<N;i++){
dp[i] = 0;
for(int j=1;;j++){
int t = (3*j-1)*j / 2;
if(t > i) break;
int tt = dp[i-t];
if(t+j <= i) tt = (tt + dp[i-t-j])%MOD;
if(j&1) dp[i] = (dp[i] + tt)%MOD;
else dp[i] = (dp[i] - tt + MOD)%MOD;
}
}
}
int Work(int n,int k) {
int ans = dp[n];
for(int i=1;;i++){
int t = k*i*(3*i-1) / 2;
if(t > n) break;
int tt = dp[n-t];
if(t + i*k <= n) tt = (tt + dp[n-t-i*k])%MOD;
if(i&1) ans = (ans - tt + MOD)%MOD;
else ans = (ans + tt)%MOD;
}
return ans;
}
int main(){
Init();
int n,k,t;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
printf("%d\n",Work(n,k));
}
return 0;
}
容斥原理dfs
题意找与n,m互质的第k个数
思路：二分
找到最小的x，使得小于或等于x的数中满足条件的数的个数大于或等于k
预处理n,m的质因数表
k是深度，也就是当前质因数位置
t是奇偶判断
s是质数乘积
n是传进去的x
void dfs(LL k,LL t,LL s,LL n) {
if(k==num) {
if(t&1) ans-=n/s;
else ans+=n/s;
return;
}
dfs(k+1,t,s,n);
dfs(k+1,t+1,s*fac[k],n); //fac[k]是质因数表
}
//二分调用
dfs(0,0,1,mid);
求（1，b）区间和（1，d）区间里面gcd(x, y) = k的数的对数（1<=x<=b , 1<= y <= d）。
b和d分别除以k之后的区间里面，只需要求gcd(x, y) = 1就可以了，这样子求出的数的对数不变。
这道题目还要求1-3 和 3-1 这种情况算成一种，因此只需要限制x
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define N 100005
typedef long long ll;
vector<int> x[N];
bool is[N];
void prime() {
memset(is, false, sizeof(is));
for (int i=0; i<N; i++) x[i].clear();
for (int j=2; j<N; j+=2) x[j].push_back(2);
for (int i=3; i<N; i+=2)
if (!is[i]) {
for (int j=i; j<N; j+=i) {
is[j] = true;
x[j].push_back(i);
}
}
}
int work(int u, int s, int w) {
int cnt = 0, v = 1;
for (int i=0; i<x[w].size(); i++) {
if ((1<<i) & s) {
cnt++;
v *= x[w][i];
}
}
int all = u/v;
if (cnt % 2 == 0) return -all;
else return all;
}
int main() {
prime();
int T, a, b, c, d, k;
scanf("%d", &T);
for (int cas=1; cas<=T; cas++) {
scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
if (k == 0) {
printf("Case %d: 0\n", cas);
continue;
}
b /= k, d /= k;
if (b > d) { a = b; b = d; d = a; }
long long ans = 0;
for (int i=1; i<=d; i++) {
k = min(i, b);
ans += k;
for (int j=1; j<(1<<x[i].size()); j++)
ans -= work(k, j, i);
}
printf("Case %d: %I64d\n", cas, ans);
}
return 0;
}
村民排队问题模型
一堆数，其中有一些两两关系，（A,B）表示A在B前面，求排列数
// UVa11174 Stand in a Line
// Rujia Liu
int mul_mod(int a, int b, int n) {
a %= n; b %= n;
return (int)((long long)a * b % n);
}
void gcd(int a, int b, int& d, int& x, int& y) {
if(!b){ d = a; x = 1; y = 0; }
else{ gcd(b, a%b, d, y, x); y -= x*(a/b); }
}
int inv(int a, int n) {
int d, x, y;
gcd(a, n, d, x, y);
return d == 1 ? (x%n+n)%n : -1;
}
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 40000 + 10;
const int MOD = 1000000007;
vector<int> sons[maxn];
int fa[maxn], fac[maxn], ifac[maxn];
int mul_mod(int a, int b) {
return mul_mod(a, b, MOD);
}
// fac[i] = (i!)%MOD, ifac[i]为fac[i]在模MOD下的逆
void preprocess() {
fac[0] = ifac[0] = 1;
for(int i = 1; i < maxn; i++) {
fac[i] = mul_mod(fac[i-1], i);
ifac[i] = inv(fac[i], MOD);
}
}
// 组合数C(n,m)除以MOD的余数
int C(int n, int m) {
return mul_mod(mul_mod(fac[n], ifac[m]), ifac[n-m]);
}
// 统计以u为根的子树有多少种排列。size为该子树的结点总数
int count(int u, int& size) {
int d = sons[u].size();
vector<int> sonsize; // 各子树的大小
size = 1;
int ans = 1;
for(int i = 0; i < d; i++) {
int sz;
ans = mul_mod(ans, count(sons[u][i], sz));
size += sz;
sonsize.push_back(sz);
}
int sz = size-1; // 非根结点的个数
for(int i = 0; i < d; i++) {
ans = mul_mod(ans, C(sz, sonsize[i]));
sz -= sonsize[i];
}
return ans;
}
int main() {
int T;
scanf("%d", &T);
preprocess();
while(T--) {
int n, m;
scanf("%d%d", &n, &m);
memset(fa, 0, sizeof(fa));
for(int i = 0; i <= n; i++) sons[i].clear();
for(int i = 0; i < m; i++) {
int a, b;
scanf("%d%d", &a, &b);
fa[a] = b;
sons[b].push_back(a);
}
// 没有父亲的结点称为虚拟结点的儿子
for(int i = 1; i <= n; i++)
if(!fa[i]) sons[0].push_back(i);
int size;
printf("%d\n", count(0, size));
}
return 0;
}
SG函数，博弈
/**********************
每组数据都改变策略
**********************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
typedef int LL;
const int MAXN = 1e4 + 5;
const int MAXM = 1e4;
int sg[MAXN];
bool Hash[MAXN];
int f[MAXN];
int N;
void getsg(int n){
memset(sg,0,sizeof sg);
for (int i=1;i<=MAXM;i++){
memset(Hash,false,sizeof Hash);
for(int j = 0; j < N && i >= f[j]; j++) {
Hash[sg[i-f[j]]] = true;
/****************上海大学校赛教训，板不要理解错。
这不是一堆拆两堆，这是每个可以转移的状态都标记。
****************/
}
for (int j=0;j<=MAXM;j++){
if (!Hash[j]){
sg[i]=j; break;
}
}
//cout << i << " " << sg[i] << endl;
}
}
int main() {
//freopen("out.txt", "w", stdout);
while(cin >> N, N) {
for(int i = 0; i < N; i++) {
scanf("%d", f + i);
}
sort(f, f + N);//一定要排序
getsg(MAXM);
int m;
cin >> m;
for(int i = 0; i < m; i++) {
int n;
cin >> n;
int ans = 0;
for(int i = 0; i < n; i++) {
int x;
scanf("%d", &x);
ans ^= sg[x];
}
printf("%s", ans ? "W" : "L");
}
printf("\n");
}
return 0;
}
/***************
独立的棋盘横向移动，看成一个子向另一个子一直在减小，NIM两子间距
***************/
/***********************
有一个操作可以把一堆拆成两堆，枚举拆分点
***********************/
for(int j = 0; j <= i - x; j++) {
Hash[sg[j] ^ sg[i - x - j]] = 1;
}
/***********************
拿走最后一个的人输，需要特判全是1的情况。
全是1，分奇偶。不全是1，同直接NIM
***********************/
/********************
两维的一样拆分成两个异或。SG[][]两维
0行，0列特殊处理。直接设置成离原点的距离。
2 2
.#
..
2 2
.#
.#
0 0
*********************/
/******************
两个操作，合并两堆，或者取掉1个
******************/
最后肯定合并成一堆再一个个取
如果全大于1，先手可以保证NIM胜利的情况下先合并
不全为1，后手可以取完一个一堆的，相当于操作了两次。
此时，DFS+记忆化搜索来解决
dp[i][j]当1的个数为i时，其他全合并起来一共j个
其中的操作包括：
把某堆只有一个的，取走
把两堆只有一个的，合并
把某堆只有一个的，合并给不是一个的
把不是一个的，取走一个
int dfs(int i, int j) {
if (dp[i][j] != -1) return dp[i][j];
if (j == 1) return dp[i][j] = dfs(i+1, 0);
dp[i][j] = 0;
if (i >= 1 && !dfs(i-1, j)) dp[i][j] = 1;
else if (j >= 1 && !dfs(i, j-1)) dp[i][j] = 1;
else if (i >= 1 && j > 0 && !dfs(i-1, j+1)) dp[i][j] = 1;
else if (i >= 2 && ((j >= 1 && !dfs(i-2, j+3)) || (j == 0 && !dfs(i-2, 2))))
dp[i][j] = 1;
return dp[i][j];
}
/****************
31 游戏
1~6各4张
****************/
直接搜索，据说记忆化也不用
反nim问题
这题与以往的博弈题的胜负条件不同，谁先走完最后一步谁输，但他也是一类Nim游戏，即为anti-nim游戏。
首先给出结论：先手胜当且仅当
①所有堆石子数都为1且游戏的SG值为0（即有偶数个孤单堆-每堆只有1个石子数）；
②存在某堆石子数大于1且游戏的SG值不为0.
字符串
Manacher算法（回文串）
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=233333;//20W
//在o(n)时间内算出以每个点为中心的最大回文串长度
int Manacher(string st){
int len = st.size();
int *p = new int[len+1];
memset(p,0,sizeof(p));
int mx = 0,id = 0;
for (int i = 1;i <= len; i++){
if (mx > i) p[i] = min(p[2*id-i],mx-i);
else p[i] = 1;
while (st[i+p[i]] == st[i-p[i]]) p[i]++;
if (i + p[i] > mx){mx = i + p[i]; id = i;}
}
int ma = 0;
for (int i = 1; i < len; i++) ma = max(ma, p[i]);
delete(p);
return ma - 1;
}
int main(){
//freopen("fuck.in","r",stdin);
char st[N];
while (~scanf("%s",st)){
string st0="$#";
for (int i=0; st[i] != '\0'; i++){
st0 += st[i]; st0 += "#";
}
printf("%d\n", Manacher(st0));
}
return 0;
}
KMP（字符串匹配）
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const int N=100007;
const int P=1000000007;
char a[N],b[N];
bool mat[N];
int Next[N];//一定要Next,next会CE
LL f[N];
void getNext(int m, char b[]){
int i = 0,j = -1;
Next[0] = -1;
while (i < m){
if (j == -1 || b[i] == b[j]){
if (b[++i] != b[++j]) Next[i]=j;
else Next[i] = Next[j];
}else j = Next[j];
}
}
//主程序里每组数据需要memset a,b数组!!!
void KMP(int n,char a[], int m, char b[]){
memset(mat, 0, sizeof(mat));
int i = 0, j = 0;
getNext(m, b);//这行视情况可以放在main里面
while (i < n && j < m){
if (j == -1 || a[i] == b[j]) i++, j++;
else j = Next[j];
if (!i && !j)break;
if (j == m){
mat[i] = 1;
//printf("mat[%d]get\n",i);
j = Next[j];
}
}
}
Tire树_刘汝佳
// LA3942 Remember the Word
// Rujia Liu
#include<cstring>
#include<vector>
using namespace std;
const int maxnode = 4000 * 100 + 10;
const int sigma_size = 26;
// 字母表为全体小写字母的Trie
struct Trie {
int ch[maxnode][sigma_size];
int val[maxnode];
int sz; // 结点总数
void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); } // 初始时只有一个根结点
int idx(char c) { return c - 'a'; } // 字符c的编号
// 插入字符串s，附加信息为v。注意v必须非0，因为0代表“本结点不是单词结点”
void insert(const char *s, int v) {
int u = 0, n = strlen(s);
for(int i = 0; i < n; i++) {
int c = idx(s[i]);
if(!ch[u][c]) { // 结点不存在
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0; // 中间结点的附加信息为0
ch[u][c] = sz++; // 新建结点
}
u = ch[u][c]; // 往下走
}
val[u] = v; // 字符串的最后一个字符的附加信息为v
}
// 找字符串s的长度不超过len的前缀
void find_prefixes(const char *s, int len, vector<int>& ans) {
int u = 0;
for(int i = 0; i < len; i++) {
if(s[i] == '\0') break;
int c = idx(s[i]);
if(!ch[u][c]) break;
u = ch[u][c];
if(val[u] != 0) ans.push_back(val[u]); // 找到一个前缀
}
}
};
#include<cstdio>
const int maxl = 300000 + 10; // 文本串最大长度
const int maxw = 4000 + 10; // 单词最大个数
const int maxwl = 100 + 10; // 每个单词最大长度
const int MOD = 20071027;
int d[maxl], len[maxw], S;
char text[maxl], word[maxwl];
Trie trie;
int main() {
int kase = 1;
while(scanf("%s%d", text, &S) == 2) {
trie.clear();
for(int i = 1; i <= S; i++) {
scanf("%s", word);
len[i] = strlen(word);
trie.insert(word, i);
}
memset(d, 0, sizeof(d));
int L = strlen(text);
d[L] = 1;
for(int i = L-1; i >= 0; i--) {
vector<int> p;
trie.find_prefixes(text+i, L-i, p);
for(int j = 0; j < p.size(); j++)
d[i] = (d[i] + d[i+len[p[j]]]) % MOD;
}
printf("Case %d: %d\n", kase++, d[0]);
}
return 0;
}
压缩Tire树
// UVa11732 strcmp() Anyone?
// Rujia Liu
#include<cstring>
#include<vector>
using namespace std;
const int maxnode = 4000 * 1000 + 10;
const int sigma_size = 26;
// 字母表为全体小写字母的Trie
struct Trie {
int head[maxnode]; // head[i]为第i个结点的左儿子编号
int next[maxnode]; // next[i]为第i个结点的右兄弟编号
char ch[maxnode]; // ch[i]为第i个结点上的字符
int tot[maxnode]; // tot[i]为第i个结点为根的子树包含的叶结点总数
int sz; // 结点总数
long long ans; // 答案
void clear() { sz = 1; tot[0] = head[0] = next[0] = 0; } // 初始时只有一个根结点
// 插入字符串s（包括最后的'\0'），沿途更新tot
void insert(const char *s) {
int u = 0, v, n = strlen(s);
tot[0]++;
for(int i = 0; i <= n; i++) {
// 找字符a[i]
bool found = false;
for(v = head[u]; v != 0; v = next[v])
if(ch[v] == s[i]) { // 找到了
found = true;
break;
}
if(!found) {
v = sz++; // 新建结点
tot[v] = 0;
ch[v] = s[i];
next[v] = head[u];
head[u] = v; // 插入到链表的首部
head[v] = 0;
}
u = v;
tot[u]++;
}
}
// 统计LCP=u的所有单词两两的比较次数之和
void dfs(int depth, int u) {
if(head[u] == 0) // 叶结点
ans += tot[u] * (tot[u] - 1) * depth;
else {
int sum = 0;
for(int v = head[u]; v != 0; v = next[v])
sum += tot[v] * (tot[u] - tot[v]); // 子树v中选一个串，其他子树中再选一个
ans += sum / 2 * (2 * depth + 1); // 除以2是每种选法统计了两次
for(int v = head[u]; v != 0; v = next[v])
dfs(depth+1, v);
}
}
// 统计
long long count() {
ans = 0;
dfs(0, 0);
return ans;
}
};
#include<cstdio>
const int maxl = 1000 + 10; // 每个单词最大长度
int n;
char word[maxl];
Trie trie;
int main() {
int kase = 1;
while(scanf("%d", &n) == 1 && n) {
trie.clear();
for(int i = 0; i < n; i++) {
scanf("%s", word);
trie.insert(word);
}
printf("Case %d: %lld\n", kase++, trie.count());
}
return 0;
}
数据结构
树状数组(逆序对)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 7;
struct binaryIndexTree{
int val[N], n;
inline void init(int n){
this->n = n;
memset(val, 0, sizeof(val));
}
inline void add(int k, int num){
for (;k <= n; k += k&-k) val[k] += num;
}
int sum(int k){
int sum = 0;
for (; k; k -= k&-k) sum += val[k];
return sum;
}
int Getsum(LL x1,LL x2){ //求任意区间和
return sum(x2) - sum(x1-1);
}
} T;
int arr[N], n;
int main(){
//freopen("in.txt", "r", stdin);
for (;~scanf("%d", &n);){
T.init(n);
int sum = 0;
for (int i = 0; i < n; i++){
scanf("%d", &arr[i]);
arr[i]++;
sum += T.sum(n) - T.sum(arr[i] - 1);
T.add(arr[i], 1);
}
int ans = sum;
for (int i = 0; i < n; i++){
sum += (n - arr[i]) - (arr[i] - 1);
ans = min(ans, sum);
}
printf("%d\n", ans);
}
}
ZKW线段树（单点修改）
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2e5 + 7;
struct segmentTree{
#define lc (t<<1)
#define rc (t<<1^1)
int sum[N], M;
inline void build(int n){
M = 1;
for (;M < n;) M <<= 1;
if (M!=1) M--;
memset(sum, sizeof(sum), 0);
for (int i = 1+M; i <= n+M; i++){
scanf("%d", &sum[i]);
}
for (int t = M; t >= 1; t--){
sum[t] = sum[lc] + sum[rc];
}
}
void add(int t, int x){
for (sum[t+=M]+=x, t>>=1; t; t>>=1){
sum[t] = sum[lc] + sum[rc];
}
}
int query(int l, int r){
int ans = 0;
for (l+=M-1,r+=M+1; l^r^1; l>>=1,r>>=1){
if (~l&1) ans += sum[l^1];
if ( r&1) ans += sum[r^1];
}
return ans;
}
} T;
ZKW线段树（RMQ区间操作）
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
double EPS = 1e-11;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, pos[N], arr[N], pre[N];
struct ZKWsegTree{
double tree[N];
int M, n;
void build(int n, double Mid){
this->n = n;
M = 1; while (M < n) M <<= 1; if (M!=1) M--;
for (int t = 1 ; t <= n; t++) tree[t+M] = 1.0*t*Mid;
for (int t = n+1; t <= M+1; t++) tree[t+M] = INF;
for (int t = M; t >= 1; t--) tree[t] = min(tree[t<<1], tree[t<<1^1]);
for (int t = 2*M+1; t >= 1; t--) tree[t] = tree[t] - tree[t>>1];
}
void update(int l, int r, double val){
double tmp;
for (l+=M-1, r+=M+1; l^r^1; l>>=1, r>>=1){
if (~l&1) tree[l^1] += val;
if ( r&1) tree[r^1] += val;
if (l > 1) tmp = min(tree[l], tree[l^1]), tree[l]-=tmp, tree[l^1]-=tmp, tree[l>>1]+=tmp;
if (r > 1) tmp = min(tree[r], tree[r^1]), tree[r]-=tmp, tree[r^1]-=tmp, tree[r>>1]+=tmp;
}
for (; l > 1; l >>= 1){
tmp = min(tree[l], tree[l^1]), tree[l]-=tmp, tree[l^1]-=tmp, tree[l>>1]+=tmp;
}
tree[1] += tree[0], tree[0] = 0;
}
double query(int l, int r){
double lAns = 0, rAns = 0;
l += M, r += M;
if (l != r){
for (; l^r^1; l>>=1, r>>=1){
lAns += tree[l], rAns += tree[r];
if (~l&1) lAns = min(lAns, tree[l^1]);
if ( r&1) rAns = min(rAns, tree[r^1]);
}
}
double ans = min(lAns + tree[l], rAns + tree[r]);
for (;l > 1;) ans += tree[l>>=1];
return ans;
}
} T;
常规线段树（区间操作区间和）
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL N = 5e5 + 7;
struct segTree{
#define lc (rt<<1)
#define rc (rt<<1^1)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1^1
LL M, sum[N], tag[N];
inline void build(LL n){
M = 1; while(M<n) M<<=1; if (M!=1) M--;
memset(tag, 0, sizeof(tag));
for (LL leaf = M+1; leaf <= n+M; leaf++) scanf("%lld", &sum[leaf]);
for (LL leaf = n+1+M; leaf <= (M<<1^1); leaf++) sum[leaf] = 0;
for (LL rt = M; rt >= 1; rt--) sum[rt] = sum[lc] + sum[rc];
}
inline void pushUp(LL rt){
sum[rt] = sum[lc] + sum[rc];
}
inline void pushDown(LL rt, LL len){
if (tag[rt] == 0) return;
tag[lc] += tag[rt];
tag[rc] += tag[rt];
sum[lc] += tag[rt] * (len>>1);
sum[rc] += tag[rt] * (len>>1);
tag[rt] = 0;
}
inline void update(LL L, LL R, LL x, LL l, LL r, LL rt){
//printf("update(%d, %d, %d, %d, %d, %d)\n", L, R, x, l, r, rt);
if (L <= l && r <= R){
tag[rt] += x;
sum[rt] += (r-l+1) * x;
return;
}
pushDown(rt, r-l+1);
LL m = (l + r) >> 1;
if (L <= m) update(L, R, x, lson);
if (m < R) update(L, R, x, rson);
pushUp(rt);
}
LL query(LL L, LL R, LL l, LL r, LL rt){
if (L <= l && r <= R) return sum[rt];
pushDown(rt, r-l+1);
LL m = (l + r) >> 1;
LL ans = 0;
if (L <= m) ans += query(L, R, lson);
if (m < R) ans += query(L, R, rson);
return ans;
}
} T;
主席树
poj2104求区间k大值
# include <cstdio>
# include <cstring>
# include <iostream>
# include <algorithm>
using namespace std;
const int N = 1e5 + 7;
int arr[N]; //arr[] 原数组的数在rank[]中的位置;
int Rank[N]; //rank[] 原数组离散化
struct ChairTree{
#define sum(x) tree[x].w
#define lson tree[rt].lc, tree[rt1].lc, l, m
#define rson tree[rt].rc, tree[rt1].rc, m+1, r
struct node{
int lc, rc, w;
node(){}
} tree[N * 20];
int root[N], cnt;
void build(){
root[0] = cnt = 0;
memset(tree, 0, sizeof(tree));
}
void add(int pos, int val, int &rt, int rt1, int l, int r){
tree[rt = ++cnt] = tree[rt1];
tree[rt].w += val;
if (l == r) return;
int m = (l + r) >> 1;
if (pos <= m) add(pos, val, lson);
else add(pos, val, rson);
}
//单点查询
int query(int k, int rt, int rt1, int l, int r){
if (l == r) return l;
int lsize = sum(tree[rt1].lc) - sum(tree[rt].lc);
int m = (l + r) >> 1;
if (lsize >= k) return query(k, lson);
else return query(k - lsize, rson);
}
//区间查询
LL query(int L, int R, int rt, int rt1, int l, int r){
if (L <= l && r <= R) return sum(rt1) - sum(rt);
if (sum(rt1) == sum(rt)) return 0;
LL ans = 0;
int m = (l + r) >> 1;
if (L <= m) ans += query(L, R, lson);
if (m < R) ans += query(L, R, rson);
return ans;
}
} T;
int main(){
//freopen("in.txt","r",stdin);
int _, l, r, k, n, q;
for (; ~scanf("%d%d", &n, &q);){
T.build();
for (int i = 1; i <= n; i++) {
scanf("%d", &arr[i]);
Rank[i] = arr[i];
}
sort(Rank + 1, Rank + n+1);//Rank存储原值
int m = unique(Rank + 1, Rank + n +1) - (Rank + 1);//这个m很重要，WA一天系列
for (int i = 1; i <= n; i++) {//离散化后的数组，仅仅用来更新
arr[i] = lower_bound(Rank + 1, Rank + m+1, arr[i]) - Rank;
}
for (int i = 1; i <= n; i++){
T.add(arr[i], 1, T.root[i], T.root[i-1], 1, m);//填m别填n
}
for (; q--;){
scanf("%d%d%d", &l, &r, &k);
int pos = T.query(k, T.root[l-1], T.root[r], 1, m);
printf("%d\n", Rank[pos]);
}
}
return 0;
}
归并树（区间第K大数）
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
const int ST_SIZE = (1 << 18) - 1;
const int MAXN = 1E5 + 5;
const int MAXM = 5005;
int N, M;
int A[MAXN]; // 要处理的数组
int I[MAXM], J[MAXM], K[MAXM]; //存查询
int nums[MAXN]; //对A排序后的数组
vector<int> dat[ST_SIZE]; //线段树节点数组
//构建线段树
// k是节点编号， 和区间[l, r)对应
void init(int k, int l, int r) {
if (r - l == 1) {
dat[k].push_back(A[l]);
} else {
int lch = k * 2 + 1, rch = 2 * k + 2;
init(lch, l, (l + r) / 2);
init(rch, (l + r) / 2, r);
dat[k].resize(r - l); //也不知道有什么用，估计是缩小空间吧
//利用STL的merge函数把两个儿子的数列合并
merge(dat[lch].begin(), dat[lch].end(), dat[rch].begin(), dat[rch].end(), dat[k].begin());
}
}
//计算[i, j)中不超过x的数的个数
//k是节点编号， 和区间[l, r)对应 (一开始 k l r 0 0 N）
int query(int i, int j, int x, int k, int l, int r) {
if (j <= l || r <= i) return 0;//完全不相交
if (i <= l && r <= j) {
return upper_bound(dat[k].begin(), dat[k].end(), x) - dat[k].begin();//完全包含
int lc = query(i, j, x, k * 2 + 1, l, (l + r) / 2);
int rc = query(i, j, x, k * 2 + 2, (l + r) / 2, r);
return lc + rc;
}
void solve() {
for(int i = 0; i < N; i++) nums[i] = A[i];
sort(nums, nums + N);
init(0, 0, N);
for (int i = 0; i < M; i++) { //[l, r)
int l = I[i] - 1, r = J[i], k = K[i];
int lb = -1, ub = N - 1; //(-1, N-1]
while(ub - lb > 1) {
int md = (ub + lb) / 2;
int ans = query(l, r, nums[md], 0, 0, N);
if (ans >= k) ub = md;
else lb = md;
}
printf("%d\n", nums[ub]);
}
}
int main(){
cin >> N >> M;
for (int i = 0; i < N; i++) scanf("%d", A + i);
for (int i = 0; i < M; i++) {
scanf("%d%d%d", I + i, J + i, K + i);
}
solve();
return 0;
}
Treap
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cassert>
using namespace std;
struct Node{
Node *ch[2];
int r, v, s;//s表示节点数
Node(int v):v(v){
ch[0]=ch[1]=NULL;
r = rand();//在cstdlib头声明
s = 1;
}
int cmp(int x){
if (x == v) return -1;
return x<v ? 0 : 1;
}
void maintain(){
s = 1;
if(ch[0]!=NULL) s+=ch[0]->s;
if(ch[1]!=NULL) s+=ch[1]->s;
}
}; //root全局使用的话可以在这里跟上*root
void rotate(Node* &o,int d){
Node *k=o->ch[d^1];
o->ch[d^1]=k->ch[d];
k->ch[d]=o;
o->maintain();
k->maintain();
o=k;
}
void insert(Node* &o,int x){//o子树中事先不存在x
if(o==NULL) o=new Node(x);
else{
//如这里改成int d=o->cmp(x);
//就不可以插入相同的值，因为d可能为-1
int d=x<(o->v)?0:1;
insert(o->ch[d],x);
if(o->ch[d]->r > o->r)
rotate(o,d^1);
}
o->maintain();
}
void remove(Node* &o,int x){
if (o==NULL) return ;//空时返回
int d=o->cmp(x);
if (d == -1){
Node *u=o;
if(o->ch[0] && o->ch[1]){
int d2=(o->ch[0]->r < o->ch[1]->r)?0:1;
rotate(o,d2);
remove(o->ch[d2],x);
}else{
if(o->ch[0]==NULL) o=o->ch[1];
else o=o->ch[0];
delete u;//这个要放里面
}
}
else remove(o->ch[d],x);
if(o) o->maintain();//之前o存在,但是删除节点后o可能就是空NULL了,所以需要先判断o是否为空
}
//返回关键字从小到大排序时的第k个值
//若返回第K大的值，只需要把ch[0]和ch[1]全互换就可以了
int kth(Node* o,int k){
assert(o && k>=1 && k<=o->s);//保证输入合法,根据实际问题返回
int s=(o->ch[0]==NULL)?0:o->ch[0]->s;
if(k==s+1) return o->v;
else if(k<=s) return kth(o->ch[0],k);
else return kth(o->ch[1],k-s-1);
}
//返回值x在树中的排名,就算x不在o树中也能返回排名
//返回值范围在[1,o->s+1]范围内
int rank(Node* o,int x){
if(o==NULL) return 1;//未找到x;
int num= o->ch[0]==NULL ? 0:o->ch[0]->s;
if(x==o->v) return num+1;
else if(x < o->v) return rank(o->ch[0],x);
else return rank(o->ch[1],x)+num+1;
}
int main(){
int n=0, v;
while(scanf("%d",&n)==1 && n){
Node *root=NULL; //初始化为NULL
for(int i=0; i<n; i++){
int x;
scanf("%d",&x);
if(root==NULL) root=new Node(x);
else insert(root,x);
}
while(scanf("%d",&v)==1){
printf("%d\n",rank(root,v));
}
}
return 0;
}
分块
分块入门 1
给出一个长为n的数列，以及n个操作，操作涉及区间加法，单点查值。
int n, blo;
int v[50005],bl[50005],atag[50005];
void add(int a, int b, int c){
for(int i=a;i<=min(bl[a]*blo,b);i++) v[i]+=c;
if(bl[a]!=bl[b])
for(int i=(bl[b]-1)*blo+1;i<=b;i++) v[i]+=c;
for(int i=bl[a]+1;i<=bl[b]-1;i++) atag[i]+=c;
}
int main(){
n=read(); blo=sqrt(n);
for(int i=1;i<=n;i++)v[i]=read();
for(int i=1;i<=n;i++)bl[i]=(i-1)/blo+1;
for(int i=1;i<=n;i++){
int f=read(),a=read(),b=read(),c=read();
if(f==0)add(a,b,c);
if(f==1)printf("%d\n",v[b]+atag[bl[b]]);
}
return 0;
}
分块入门 2
给出一个长为n的数列，以及n个操作，操作涉及区间加法，询问区间内小于某个值x的元素个数。
int n,blo;
int v[50005], bl[50005], atag[50005];
vector<int>ve[505];
void reset(int x){
ve[x].clear();
for(int i=(x-1)*blo+1;i<=min(x*blo,n);i++)
ve[x].push_back(v[i]);
sort(ve[x].begin(),ve[x].end());
}
void add(int a,int b,int c){
for(int i=a;i<=min(bl[a]*blo,b);i++)
v[i]+=c;
reset(bl[a]);
if(bl[a]!=bl[b]){
for(int i=(bl[b]-1)*blo+1;i<=b;i++)v[i]+=c;
reset(bl[b]);
}
for(int i=bl[a]+1;i<=bl[b]-1;i++)
atag[i]+=c;
}
int query(int a,int b,int c){
int ans=0;
for(int i=a;i<=min(bl[a]*blo,b);i++)
if(v[i]+atag[bl[a]]<c)ans++;
if(bl[a]!=bl[b])
for(int i=(bl[b]-1)*blo+1;i<=b;i++)
if(v[i]+atag[bl[b]]<c)ans++;
for(int i=bl[a]+1;i<=bl[b]-1;i++){
int x=c-atag[i];
ans+=lower_bound(ve[i].begin(),ve[i].end(),x)-ve[i].begin();
}
return ans;
}
int main(){
n=read();blo=sqrt(n);
for(int i=1;i<=n;i++)v[i]=read();
for(int i=1;i<=n;i++){
bl[i]=(i-1)/blo+1;
ve[bl[i]].push_back(v[i]);
}
for(int i=1;i<=bl[n];i++)
sort(ve[i].begin(),ve[i].end());
for(int i=1;i<=n;i++){
int f=read(),a=read(),b=read(),c=read();
if(f==0)add(a,b,c);
if(f==1)printf("%d\n",query(a,b,c*c));
}
return 0;
}
分块入门 3
给出一个长为n的数列，以及n个操作，操作涉及区间加法，询问区间内小于某个值x的前驱（比其小的最大元素）。
int n,blo;
int v[100005],bl[100005],atag[100005];
set<int>st[105];
void add(int a,int b,int c){
for(int i=a;i<=min(bl[a]*blo,b);i++){
st[bl[a]].erase(v[i]);
v[i] += c;
st[bl[a]].insert(v[i]);
}
if(bl[a]!=bl[b]){
for(int i=(bl[b]-1)*blo+1;i<=b;i++){
st[bl[b]].erase(v[i]);
v[i] += c;
st[bl[b]].insert(v[i]);
}
}
for(int i=bl[a]+1;i<=bl[b]-1;i++)
atag[i]+=c;
}
int query(int a,int b,int c){
int ans=-1;
for(int i=a;i<=min(bl[a]*blo,b);i++){
int val=v[i]+atag[bl[a]];
if(val<c)ans=max(val,ans);
}
if(bl[a]!=bl[b])
for(int i=(bl[b]-1)*blo+1;i<=b;i++){
int val=v[i]+atag[bl[b]];
if(val<c)ans=max(val,ans);
}
for(int i=bl[a]+1;i<=bl[b]-1;i++){
int x=c-atag[i];
set<int>::iterator it=st[i].lower_bound(x);
if(it==st[i].begin())continue;
--it;
ans=max(ans,*it+atag[i]);
}
return ans;
}
int main(){
n=read();blo=1000;
for(int i=1;i<=n;i++)v[i]=read();
for(int i=1;i<=n;i++){
bl[i]=(i-1)/blo+1;
st[bl[i]].insert(v[i]);
}
for(int i=1;i<=n;i++){
int f=read(),a=read(),b=read(),c=read();
if(f==0)add(a,b,c);
if(f==1)printf("%d\n",query(a,b,c));
}
return 0;
}
左偏树
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int MAXN = 1e5 + 5;
struct node{
int l,r,dis,key;
} tree[MAXN];
int far[MAXN];
int Find(int x) {
if(far[x] == x) return x;
return far[x] = Find(far[x]);
}
int merge(int a,int b){
if(!a) return b;
if(!b) return a;
if(tree[a].key < tree[b].key) swap(a, b);//大堆
tree[a].r = merge(tree[a].r,b);
far[tree[a].r] = a;//并查
if(tree[tree[a].l].dis < tree[tree[a].r].dis) swap(tree[a].l,tree[a].r);
if(tree[a].r)tree[a].dis = tree[tree[a].r].dis + 1;
else tree[a].dis = 0;
return a;
}
int pop(int a){
int l = tree[a].l;
int r = tree[a].r;
far[l] = l;//因为要暂时删掉根，所以左右子树先作为根
far[r] = r;
tree[a].l = tree[a].r = tree[a].dis = 0;
return merge(l,r);
}
int main(){
int N, M;
while(cin >> N){
for(int i = 1; i <= N; i++){
int x;
far[i] = i;
scanf("%d", &x);
tree[i].key = x;
tree[i].l = tree[i].r = tree[i].dis = 0;
}
cin >> M;
while(M--) {
int x, y;
scanf("%d%d", &x, &y);
x = Find(x);
y = Find(y);
if(x == y) {
printf("-1\n");
} else {
int ra = pop(x);
tree[x].key /= 2;
ra = merge(ra, x);
int rb = pop(y);
tree[y].key /= 2;
rb = merge(rb, y);
x = merge(ra, rb);
printf("%d\n", tree[x].key);
}
}
}
return 0;
}
Splay
#include<cstdio>
#include<algorithm>
using namespace std;
struct Node{
int key;//size
Node *l,*r,*f;//left,right,father
};
class SplayTree{
public:
void Init(){rt=NULL;}
void Zag(Node *x){//left rotate
Node *y=x->f;//y is the father of x
y->r = x->l;
if (x->l)x->l->f = y;//if x has left child
x->f =y->f;
if (y->f){//y is not root
if (y==y->f->l)y->f->l=x;//y if left child
else y->f->r=x;//y is right child
}
y->f=x; x->l=y;
}
void Zig(Node *x){//right rotate
Node *y=x->f;//y is the father of x
y->l = x->r;
if (x->r)x->r->f=y;
x->f = y->f;
if (y->f){
if (y==y->f->l)y->f->l=x;
else y->f->r=x;
}
y->f=x;
x->r=y;
}
void Splay(Node *x){
while (x->f){
Node *p=x->f;
if (!p->f){
if (x==p->l)Zig(x);
else Zag(x);
}else if (x==p->l){
if (p==p->f->l){Zig(p);Zig(x);}
else {Zig(x);Zag(x);}
}else {//x==p->r
if (p==p->f->r){Zag(p);Zag(x);}
else {Zag(x);Zig(x);}
}
}
rt=x;
}
Node *Find(int x){
Node *T=rt;
while (T){
if (T->key==x){Splay(T);return T;}
else if (x<T->key)T=T->l;
else T=T->r;
}
return T;
}
void Insert(int x){
Node *T=rt,*fa=NULL;
while (T){
fa=T;
if (x<T->key)T=T->l;
else if(x>T->key)T=T->r;
else return ;//two the same keys
}
T=(Node*)malloc(sizeof(Node));
T->key=x;
T->l=T->r=NULL;
T->f=fa;
if (fa){
if (fa->key>x)fa->l=T;
else fa->r=T;
}
Splay(T);
}
void Delete(int x){
Node *T=Find(x);
if (NULL==T)return ;//error
rt=Join(T->l,T->r);
}
Node *Maxnum(Node *t){
Node *T=t;
while (T->r)T=T->r;
Splay(T);
return T;
}
Node *Minnum(Node *t){
Node *T=t;
while (T->l)T=T->l;
Splay(T);
return T;
}
Node *Last(int x){
Node *T=Find(x);
T=T->l;
return (Maxnum(T));
}
Node *Next(int x){
Node *T=Find(x);
T=T->r;
return (Minnum(T));
}
Node *Join(Node *t1,Node *t2){
if (NULL==t1)return t2;
if (NULL==t2)return t1;
Node *T=Maxnum(t1);
T->l=t2;
return T;
}
void Split(int x,Node *&t1,Node *&t2){
Node *T=Find(x);
t1=T->l; t2=T->r;
}
void Inorder(Node *T){
if (NULL==T)return ;
Inorder(T->l);
printf("%d->",T->key);
Inorder(T->r);
}
void _Delete(){Delete(rt);}
void Delete(Node *T){
if (NULL==T)return ;
Delete(T->l);
Delete(T->r);
free(T);
}
private:
Node *rt;//root
};
AVL树
//codevs1285 莯ʕѸ˹
//by cww97
#include<cstdio>
#include<iostream>
#include<algorithm>
#define INF 0xfffffff
#define BASE 1000000
using namespace std;
int ans=0;
struct Node{
int x,bf,h;//bf=balance factor,h=height
Node *l,*r;
};
class AVLTree{
public:
void Init() { rt = NULL; }
int H(Node *T){return (T==NULL)?0:T->h;}
int BF(Node *l,Node *r){//get balance factor
if (NULL==l && NULL==r) return 0;
else if (NULL == l) return -r->h;
else if (NULL == r) return l->h;
return l->h - r->h;
}
Node *Lrorate(Node *a){//left rorate
Node *b;
b=a->r;
a->r=b->l;
b->l=a;
a->h=max(H(a->l),H(a->r)) + 1;
b->h=max(H(b->l),H(b->r)) + 1;
a->bf=BF(a->l,a->r);
b->bf=BF(b->l,b->r);
return b;
}
Node *Rrorate(Node *a){//right rorate
Node *b;
b=a->l;
a->l=b->r;
b->r=a;
a->h=max(H(a->l),H(a->r)) + 1;
b->h=max(H(b->l),H(b->r)) + 1;
a->bf=BF(a->l,a->r);
b->bf=BF(b->l,b->r);
return b;
}
Node *LRrorate(Node *a){//left then right
a->l = Lrorate(a->l);
Node *c;
c=Rrorate(a);
return c;
}
Node *RLrorate(Node *a){//right then left
a->r=Rrorate(a->r);
Node *c;
c=Lrorate(a);
return c;
}
void Insert(int x){_Insert(rt,x);}
void _Insert (Node *&T,int x){
if (NULL==T){
T=(Node*)malloc(sizeof(Node));
T->x=x;
T->bf=0;T->h=1;
T->l=T->r=NULL;
return ;
}
if (x < T->x) _Insert(T->l,x);
else if (x > T->x) _Insert(T->r,x);
else return ; //error :the same y
T->h=max(H(T->l),H(T->r))+1;//maintain
T->bf=BF(T->l,T->r);
if (T->bf > 1 || T->bf < -1){//not balanced
if (T->bf > 0 && T->l->bf > 0)T=Rrorate(T);
else if (T->bf < 0 && T->r->bf < 0)T=Lrorate(T);
else if (T->bf > 0 && T->l->bf < 0)T=LRrorate(T);
else if (T->bf < 0 && T->r->bf > 0)T=RLrorate(T);
}
}
void GetPet(int x){//get pet or person
if (NULL==rt){return ;}
int small=0,large=INF;
//printf("x=%d\n",x);
int flag;
if (Find(rt,x,small,large)){
printf("find %d\n",x);
_Delete(rt,x);
}else if (small==0)flag=1;
else if (large==INF)flag=0;
else if (large-x<x-small)flag=1;
else flag=0;
if (!flag){//choose large
_Delete(rt,small);
ans=(ans+x-small)%BASE;
}else {
_Delete(rt,large);
ans=(ans+large-x)%BASE;
}
}
bool Find(Node *T,int x,int &small,int &large){
if (NULL==T)return 0;
if (x==T->x)return 1;
if (x<T->x){
large=min(large,T->x);
return Find(T->l,x,small,large);
}else{
small=max(small,T->x);
return Find(T->r,x,small,large);
}
}
void _Delete(Node *&T,int x){
if (NULL==T)return ;
if (x < T->x){//y at left
_Delete(T->l,x);
T->bf=BF(T->l,T->r);
if (T->bf<-1){
if (1==T->r->bf)T=RLrorate(T);
else T=Lrorate(T);//bf==0 or -1
}
}else if (x > T->x){//y at right
_Delete(T->r,x);
T->bf=BF(T->l,T->r);
if (T->bf>1){
if (-1==T->l->bf)T=LRrorate(T);
else T=Rrorate(T);//bf==0 or 1
}
}else {//here is x
if (T->l&&T->r){//left &&right
Node *t=T->l;
while (t->r)t=t->r;
T->x=t->x;
_Delete(T->l,t->x);
T->bf=BF(T->l,T->r);
if (T->bf<-1){
if (1==T->r->bf)T=RLrorate(T);
else T=Lrorate(T);//bf==0 or -1
}
}else {//left || right
Node *t=T;
if (T->l)T=T->l;
else if(T->r)T=T->r;
else {free(T);T=NULL;}
if (T)free(t);
}
}
}
//Debug,you will not need it at this problem
void show(){InOrder(rt);puts("EndShow");}
void InOrder(Node *T){//print l rt r
if (NULL==T)return ;
InOrder(T->l);
printf("%d ",T->x);
InOrder(T->r);
}
void Free(){FreeTree(rt);}
void FreeTree(Node *T){
if (NULL==T)return ;
FreeTree(T->l);
FreeTree(T->r);
free(T);
}
private:
Node *rt;//root
};
int main(){
freopen("fuck.in","r",stdin);
int n,x,op,a=0,b=0;
scanf("%d",&n);
AVLTree T; T.Init();
for (;n--;){
scanf("%d%d",&op,&x);
//if pets>people put pets into the tree
//else put people into the tree
if (op==0){//come a pet
a++;
if (a>b)T.Insert(x);//more pet
else T.GetPet(x);//more people
}else{//come a person
b++;
if (a<b)T.Insert(x);//more people
else T.GetPet(x);//more pet
}
}
printf("%d\n",ans%BASE);
T.Free();
return 0;
}
图论
最小生成树（prim）
hdu1102
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=107;
int n,g[N][N];
int prim(){
int minw[N];//MinWeight
bool used[N];
memset(used,0,sizeof(used));
memset(minw,0x7f,sizeof(minw));
minw[1]=0;
int sum=0;
while (1){
int v=-1;
for (int i=1;i<=n;i++){
if (!used[i]&&(v==-1||minw[i]<minw[v]))v=i;
}
if (v==-1)break;
used[v]=1;
sum+=minw[v];
for (int i=0;i<=n;i++){
minw[i]=min(minw[i],g[v][i]);
}
}
return sum;
}
int main(){
for (;scanf("%d",&n)==1;){
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++) scanf("%d",&g[i][j]);
int x,y,q;
scanf("%d",&q);
for (;q--;){
scanf("%d%d",&x,&y);
g[x][y]=g[y][x]=0;
}
printf("%d\n",prim());
}
return 0;
}
次小生成树
#include<iostream>
#include<cstdio>
#include<cstring>
#include<climits>
#include<algorithm>
using namespace std;
#define N 510
int map[N][N], lowcost[N], pre[N], max1[N][N], stack[N];
bool visit[N];
int n, m, sum;
void prim(){ //默认1在MST中
int temp, k;
int top; //保存最小生成树的结点
memset(visit, false, sizeof(visit)); //初始化
visit[1] = true;
sum = top = 0;
for(int i = 1; i <= n; ++i){
pre[i] = 1;
lowcost[i] = map[1][i];
}
lowcost[1] = 0;
stack[top++] = 1; //保存MST的结点
for(int i = 1; i <= n; ++i){
temp = INT_MAX;
for(int j = 1; j <= n; ++j)
if(!visit[j] && temp > lowcost[j])
temp = lowcost[k = j];
if(temp == INT_MAX) break;
visit[k] = true;
sum += temp;
for(int j = 0; j < top; ++j) //新加入点到MST各点路径最大值
max1[stack[j]][k] = max1[k][stack[j]] = max(max1[stack[j]][pre[k]], temp);
stack[top++] = k; //保存MST的结点
for(int j = 1; j <= n; ++j) //更新
if(!visit[j] && lowcost[j] > map[k][j]){
lowcost[j] = map[k][j];
pre[j] = k; //记录直接前驱
}
}
}
int main(){
int ncase, start, end, cost, minn;
scanf("%d", &ncase);
while(ncase--) {
for(int i = 1; i < N; ++i) //初始化不为0,1必须用循环。。。。
for(int j = 1; j < N; ++j){
map[i][j] = INT_MAX;
max1[i][j] = 0;
}
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i){
scanf("%d%d%d", &start, &end, &cost);
//if(cost < map[start][end])(POJ竟然出现重边的时候不选择最小的~~~)
map[start][end] = map[end][start] = cost;
}
prim();
minn = INT_MAX;
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
if(i != j && i != pre[j] && j != pre[i]) //枚举MST以外的边
minn = min(minn, map[i][j] - max1[i][j]); //求出{MST外加入边-MST环上权值最大边}最小值
if(minn != 0) printf("No\n");
else printf("Yes\n");
}
return 0;
}
最短路（SPFA）
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100007;
const int INF=0x3f3f3f3f;
struct SPFA{
struct Edge{
int from,to,cost,a,b;
Edge(){}
Edge(int x,int y,int z,int a,int b):
from(x),to(y),cost(z),a(a),b(b){}
bool canPass(){return a>=cost;}//题目里的
void read(){scanf("%d%d%d%d%d",&from,&to,&a,&b,&cost);}
};
int n, m;
vector <Edge> edges;
vector <int > G[N];
inline void AddEdge(){
Edge e; e.read();
if (!e.canPass())return ;
edges.push_back(e);
int top = edges.size();
G[e.from].push_back(top-1);
}
inline void Init(int n,int m){
this -> n = n; this -> m = m;
edges.clear();
for (int i=0;i<=n;i++)G[i].clear();
for (int i=1;i<=m;i++) AddEdge();
}
bool inq[N];
int d[N],cnt[N];
inline int spfa(int s,int t){
queue<int> Q;
memset(inq, 0, sizeof(inq));
memset(cnt, 0, sizeof(cnt));
memset( d ,INF,sizeof( d ));
d[s] = 0;inq[s]=1;Q.push(s);
for (;!Q.empty();){
int u = Q.front();Q.pop();
inq[u]= 0;
for (int i=0;i<G[u].size();i++){
Edge &e = edges[G[u][i]];
int val = d[u],s = d[u] ;//
val %= (e.a + e.b);//特殊题目
if (val>e.a) s+=e.b-(val-e.a);//
else if (e.a<val+e.cost)s+=e.a+e.b-val;//
if (d[u]<INF&&s+e.cost<d[e.to]){
d[e.to] = s + e.cost;
if (!inq[e.to]){
Q.push(e.to);inq[e.to] = 1;
if (++cnt[e.to]>n)return INF;
}
}
}
}
return d[t];
}
}g;
int main(){
//freopen("in.txt","r",stdin);
int n,m,s,t;
for (int T=0;~scanf("%d%d%d%d",&n,&m,&s,&t);){
g.Init(n,m,s,t);
int ans = g.spfa(s,t);
printf("Case %d: %d\n",++T,ans);
}
return 0;
}
找负环
bool find_negative_loop() {
memset(d, 0, sizeof(d));
for(int i = 0; i < V; i++) {
for(int j = 0; j < E; j++) {
edge e = es[j];
if(d[e.to] > d[e.form] + e.cost) {
d[e.to] = d[e.from] + e.cost;
if(i == V - 1) return true;
}
}
}
return false;
}
最短路Dijkstra同时求解次短路
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 7;
struct Edge{
int to;
LL cost;
Edge(int tv = 0, LL tc = 0):to(tv), cost(tc){}
};
vector<Edge> G[N];
int n, m;
LL dist[N]; //最短距离
LL dist2[N]; //次短距离
LL Dijkstra(){
memset(dist, INF, sizeof(dist));
memset(dist2, INF, sizeof(dist2));
//从小到大的优先队列
//使用pair而不用edge结构体
//是因为这样我们不需要重载运算符
//pair是以first为主关键字进行排序
priority_queue<P, vector<P>, greater<P> > Q;
//初始化源点信息
dist[1] = 0;
Q.push(P(0, 1));
//同时求解最短路和次短路
for(; !Q.empty();){
P p = Q.top(); Q.pop();
//first为s->to的距离，second为edge结构体的to
int v = p.second;
LL d = p.first;
//当取出的值不是当前最短距离或次短距离，就舍弃他
if (dist2[v] < d) continue;
for (int i = 0; i < G[v].size(); i++){
Edge &e = G[v][i];
LL d2 = d + e.cost;
if (dist[e.to] > d2){
swap(dist[e.to], d2);
Q.push(P(dist[e.to], e.to));
}
//printf("dist2[%d] = %d, d2 = %d\n", e.to, dist2[e.to], d2);
if (dist2[e.to] > d2 && dist[v] < d2){
dist2[e.to] = d2;
Q.push(P(dist2[e.to], e.to));
}
}
}
return dist2[n];
}
多源最短路(Floyed)
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 200;
int n, G[N][N], d[N][N];
inline void Init(int n){
this -> n = n;
memset(G,INF,sizeof(G));
memset(d,INF,sizeof(d));
}
inline void AddEdge(int f,int t){
G[f][t] = d[f][t] = 1;
}
inline void floyed(){
for (int k=1;k<=n;k++)
for (int i=1;i<=n;i++)if (i!=k)
for (int j=1;j<=n;j++)if (j!=i&&j!=k)
d[i][j]=min(d[i][j], d[i][k]+d[k][j]);
}
欧拉回路dfs
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long LL;
const int N = 9999,INF=0x3f3f3f3f;
struct EulerCircle{
struct Edge{
int to,nxt;
Edge(){}
Edge(int x,int y):to(x),nxt(y){}
}edges[N];
int head[N],n,E;
inline void Init(const int &n){
this->n = n; E = 0;
for (int i=0;i<=n;i++)head[i]=-1;
}
inline void AddEdge(int f,int t){
edges[++E] = Edge(t,head[f]);
head[f] = E ;
}
stack<int >S;
bool vis[N];//use when dfs
inline void dfs(int x){//get EulerCircle
Edge e;
for (int i=head[x];i!=-1;i=e.nxt){
e = edges[i];
if (vis[i])continue;
vis[i] = 1;
dfs(e.to);
S.push(x);
}
}
inline void getEulerCircle(){
while (!S.empty())S.pop();
memset(vis,0,sizeof(vis));
dfs(1);
for (;!S.empty();S.pop())
printf("%d ",S.top());
puts("1");
}
} g ;
混合图欧拉回路
解析见紫书376
ATTENTION:需要注意的是，网络流里是有反向边的，dinic跑完之后反向边不要添加到新图里面了
加到Dinic里面
inline void buildEuler(int n){
for (int i=1;i<=n;i++){
for (int nxt,j=head[i];j!=-1;j=nxt){
Edge &e = edges[j]; nxt = e.nxt;
if (e.to==s||e.to==t) continue ;
if (!e.cap)continue;
if (e.flow==e.cap)gg.AddEdge(e.to,e.from);
else gg.AddEdge(e.from, e.to);
}
}
}
main哇哦
int d[N];//degree = out - in
bool work(int n){
int flow = 0;
for (int i=1;i<=n;i++){
if (d[1]&1)return 0;
if (d[i]>0){
g.AddEdge(g.s,i,d[i]>>1);
flow += d[i]>>1;
}else if (d[i]<0)
g.AddEdge(i,g.t,-(d[i]>>1));
}
if (flow != g.maxFlow()) return 0;
return 1;
}
int main(){
//freopen("in.txt","r",stdin);
int T,x,y,n,m;
scanf("%d",&T);
for (char ch;T--;){
scanf("%d%d",&n,&m);
g.Init(n,0,n+1);
gg.Init(n);
memset(d,0,sizeof(d));
for (int i=1;i<=m;i++){
scanf("%d%d %c\n",&x,&y,&ch);
if (ch=='D') gg.AddEdge(x,y);
else g.AddEdge(x,y,1);
d[x]++;d[y]--;//Degree
}
if (!work(n))puts("No euler circuit exist");
else {
g.buildEuler(n);
gg.getEulerCircle();
}
if (T)puts("");
}
return 0;
}
拓扑排序topsort
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
const int maxn=107;
int x,y,m,n;
bool f[maxn][maxn];
int in[maxn];
int main(){
//freopen("fuck.in" ,"r",stdin);
while(scanf("%d%d",&n,&m)==2&&(m||n)){
memset(f ,0,sizeof(f ));
memset(in,0,sizeof(in));
for(;m--;){
scanf("%d%d",&x,&y);
f[x][y] = 1;
in[y]++;
}
int A=0,ans[maxn];
queue<int>Q;
for (int i=1;i<=n;i++)if (in[i]==0)Q.push(i);
while(!Q.empty()){
int x=Q.front(); Q.pop();
ans[++A]=x;
for(int j=1;j<=n;j++)if (f[x][j]){
if (--in[j]==0)Q.push(j);
}
}
for (int i=1;i<A;i++)printf("%d ",ans[i]);
printf("%d\n",ans[A]);
}
return 0;
}
最大流（Dinic）
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
const int N = 9999;
struct Dinic{
struct Edge{
int from,to,cap,flow,nxt;
Edge(){}
Edge(int u,int v,int c,int f,int n):
from(u),to(v),cap(c),flow(f),nxt(n){}
}edges[N];
int n, s, t, E, head[N];
bool vis[N]; //use when bfs
int d[N],cur[N];//dist,now edge,use in dfs
inline void AddEdge(int f,int t,int c){
edges[++E] = Edge(f,t,c,0,head[f]);
head[f] = E;
edges[++E] = Edge(t,f,0,0,head[t]);
head[t] = E;
}
inline void Init(int n,int s,int t){
this -> n = n ; E = -1;
this -> s = s ; head[s] = -1;
this -> t = t ; head[t] = -1;
for (int i=0;i<=n;i++) head[i] = -1;
}
inline bool BFS(){
memset(vis,0,sizeof(vis));
queue<int >Q;
d[s] = 0; vis[s] = 1;
for (Q.push(s);!Q.empty();){
int x = Q.front(); Q.pop();
for (int nxt,i = head[x];i!=-1;i = nxt){
Edge &e = edges[i]; nxt = e.nxt;
if (vis[e.to]||e.cap<=e.flow)continue;
vis[e.to]=1;
d[e.to]=d[x]+1;
Q.push(e.to);
}
}
return vis[t];
}
inline int DFS(const int& x,int a){
if (x==t||a==0){return a;}
int flow = 0, f, nxt;
for (int& i=cur[x];i!=-1;i=nxt){
Edge& e = edges[i]; nxt = e.nxt;
if (d[x]+1!=d[e.to])continue;
if ((f=DFS(e.to,min(a,e.cap-e.flow)))<=0)continue;
e.flow += f;
edges[i^1].flow-=f;//ϲߍ
flow+=f; a-=f;
if (!a) break;
}
return flow;
}
inline int maxFlow(){return maxFlow(s,t);}
inline int maxFlow(int s, int t){
int flow = 0;
for (;BFS();){
for (int i=0;i<=n;i++)cur[i]=head[i];
flow += DFS(s,INF) ;
}
return flow;
}
inline void buildEuler(int n){
for (int i=1;i<=n;i++){
for (int nxt,j=head[i];j!=-1;j=nxt){
Edge &e = edges[j]; nxt = e.nxt;
if (e.to==s||e.to==t) continue ;
if (!e.cap)continue;
if (e.flow==e.cap)gg.AddEdge(e.to,e.from);
else gg.AddEdge(e.from, e.to);
}
}
}
} g ;
费用流（SPFA）
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = 2007;
const int INF=0x3f3f3f3f;
const double EPS = 1e-6;
struct MCMF{
struct Edge{
int from,to,cap,flow,nxt;
double cost;
Edge(){}
Edge(int x,int y,int z,int u,double v,int n){
from=x;to=y;cap=z;flow=u;cost=v;nxt=n;
}
}edges[N];
int E,head[N];
int n,s,t,inq[N],p[N],a[N];
double d[N];
inline void Init(int n,int s,int t){
this->n = n; E = -1;
this->s = s; this->t = t;
memset(head,-1,sizeof(head));
}
inline void AddEdge(int f,int t,int c,double w){
edges[++E] = Edge(f,t,c,0, w,head[f]);
head[f] = E;
edges[++E] = Edge(t,f,0,0,-w,head[t]);
head[t] = E;
}
bool spfa(int s,int t,int flow,double &cost){
for (int i=0;i<=n;i++)d[i]=INF;
memset(inq,0,sizeof(inq));
d[s]=0;inq[s]=1;p[s]=0;a[s]=INF;
queue<int>Q;Q.push(s);
for (;!Q.empty();){
int nxt, u =Q.front();Q.pop();inq[u]=0;
for (int i=head[u];i!=-1;i=nxt){
Edge &e = edges[i]; nxt = e.nxt;
if (e.cap<=e.flow||d[e.to]<=d[u]+e.cost)continue;
d[e.to] = d[u] + e.cost;
p[e.to] = i;
a[e.to] = min(a[u],e.cap-e.flow);
if (!inq[e.to]){Q.push(e.to);inq[e.to]=1;}
}
}
if (d[t]==INF)return 0;//false
flow += a[t];
cost += (double)d[t]*(double)a[t];
//printf("cost=%.2lf\n",cost);
for (int u=t;u!=s;u=edges[p[u]].from){
edges[p[u] ].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
return 1;//true
}
//需要保证初始网络中没有负权
double mcmf(){
int flow =0;
double cost = 0;
for (;spfa(s,t,flow,cost););
return cost;
}//MinCostMaxFlow
} g ;
强连通分量Tarjan
hdu5934
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL N = 1e3 + 7;
const LL INF = 1e8 + 7;
struct bomb{
LL x, y, r, c;
void read(){
scanf("%lld%lld%lld%lld", &x, &y, &r, &c);
}
} bombs[N];
LL sqr(LL x){return x*x;}
double dist(LL i, LL j){
return sqrt(sqr(bombs[i].x - bombs[j].x)
+sqr(bombs[i].y - bombs[j].y));
}
LL n;
struct tarjan //复杂度O(N+M)
{
const static LL MAXN = N; //点数
const static LL MAXM = N*N;//边数
struct Edge{
LL to,next;
} edge[MAXM];
LL head[MAXN], tot;
LL Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc
LL Index,top;
LL scc;//强连通分量的个数
LL num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc //num数组不一定需要，结合实际情况
bool Instack[MAXN];
void addedge(LL u,LL v){
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void Tarjan(LL u){
LL v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(LL i = head[u]; ~i; i = edge[i].next){
v = edge[i].to;
if( !DFN[v] ){
Tarjan(v);
Low[u] = min(Low[u], Low[v]);
}
else if(Instack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(Low[u] == DFN[u]){
scc++;
do{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc;
num[scc]++;
}while( v != u);
}
}
LL in[MAXN];
LL solve(LL N){
memset(DFN, 0, sizeof(DFN));
memset(num, 0, sizeof(num));
memset(Instack, 0, sizeof(Instack));
Index = scc = top = 0;
for(LL i = 1; i <= N; i++) if(!DFN[i]) Tarjan(i);
//for this problem
memset(in, 0, sizeof(in));
for (LL u = 1; u <= n; u++){
for (LL e = head[u]; ~e; e = edge[e].next){
LL v = edge[e].to;
if (Belong[u] != Belong[v]) in[Belong[v]]++;
}
}
LL ans = 0;
for (LL i = 1; i <= scc; i++) if (in[i] == 0){
LL cost = INF;
for (LL j = 1; j <= n; j++) if (Belong[j] == i){
cost = min(cost, bombs[j].c);
}
ans += cost;
}
return ans;
}
void init(){
tot = 0;
memset(head, -1, sizeof(head));
}
} g;
int main(){
//freopen("in.txt", "r", stdin);
LL T;
scanf("%lld", &T);
for (LL cas = 1; cas <= T; cas++){
printf("Case #%lld: ", cas);
scanf("%lld", &n);
for (LL i = 1; i <= n; i++){
bombs[i].read();
}
g.init();
for (LL i = 1; i < n; i++){
for (LL j = i+1; j <= n; j++){
double d = dist(i, j);
if (bombs[i].r >= d) g.addedge(i, j);
if (bombs[j].r >= d) g.addedge(j, i);
}
}
LL ans = g.solve(n);
printf("%lld\n", ans);
}
return 0;
}
倍增LCA + 最大生成树
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int N = 1e5 + 5;
int n,m;
struct gragh{
struct Edge{
int from,to,w;
Edge(){}
Edge(int x,int y,int z):from(x),to(y),w(z){}
bool operator < (const Edge& a)const{
return w < a.w;
}
}edges[N],be[N];
int E,f[N],fa[N][20],di[N][20],dep[N];
bool vis[N];
vector<int >G[N];
int F(int x){//鼯
return f[x]==x?x:(f[x]=F(f[x]));
}
inline void link(int x,int y,int z){
edges[++E]=Edge(x,y,z);
G[x].push_back(E);
}
void build(){
E=0;
for (int i=1;i<=n;i++)G[i].clear();
int x,y,z;
for (int i=1;i<=n;i++)f[i]=i;
for (int i=1;i<=m;i++){
scanf("%d%d%d",&x,&y,&z);
be[i]=Edge(x,y,z);
f[F(x)]=F(y);
}
}
void kruskal(){
int treenum = 0;//forests
memset(vis,0,sizeof(vis));
for (int i=1;i<=n;i++)if (!vis[F(i)]){
treenum++;vis[F(i)]=1;
}
for (int i=1;i<=n;i++)f[i]=i;
sort(be+1,be+m+1);
int cnt = 0;
for (int i=m;i>=1;i--){
int x = be[i].from;
int y = be[i].to ;
if (F(x)==F(y))continue;
f[F(x)]=F(y);
cnt++;
link(x,y,be[i].w);
link(y,x,be[i].w);
if (cnt==n-treenum)break;
}
}
void dfs(int x){
vis[x] = 1;
for (int i=1;i<=17;i++){
if(dep[x]<(1<<i))break;
fa[x][i]=fa[fa[x][i-1]][i-1];
di[x][i]=min(di[x][i-1],di[fa[x][i-1]][i-1]);
}
for (int i=0;i<G[x].size();i++){
Edge e = edges[G[x][i]];
if (vis[e.to])continue;
fa[e.to][0] = x;
di[e.to][0] = e.w;
dep[e.to] = dep[x]+1;
dfs(e.to);
}
}
int lca(int x,int y){
if (dep[x]<dep[y])swap(x,y);
int t = dep[x] - dep[y];
for (int i=0;i<=17;i++)
if ((1<<i)&t) x = fa[x][i];
for (int i=17;i>=0;i--)
if (fa[x][i]!=fa[y][i]){
x=fa[x][i];y=fa[y][i];
}
if (x==y)return x;
return fa[x][0];
}
int ask(int x,int f){//f:father
int ans = INF;
int t = dep[x]-dep[f];
for (int i=0;i<=17;i++)if(t&(1<<i)){
ans=min(ans,di[x][i]);
x = fa[x][i];
}
return ans;
}
void work(){
build();
kruskal();
memset(vis,0,sizeof(vis));
for (int i=1;i<=n;i++)if(!vis[i])dfs(i);
int q,x,y;
scanf("%d",&q);
while (q--){
scanf("%d%d",&x,&y);
if (F(x)!=F(y))puts("-1");
else {
int t = lca(x,y);
x = ask(x,t);
y = ask(y,t);
printf("%d\n",min(x,y));
}
}
}
}g;
int main(){
//freopen("truck.in","r",stdin);
//freopen("truck.out","w",stdout);
for (;~scanf("%d%d",&n,&m);)g.work();
return 0;
}
树链剖分
struct TreeChain{
struct Edge{
int from, to, nxt;
Edge(){}
Edge(int u, int v, int n):
from(u), to(v), nxt(n){}
}edges[N];
int n, E, head[N];
int tim;
int siz[N]; //用来保存以x为根的子树节点个数
int top[N]; //用来保存当前节点的所在链的顶端节点
int son[N]; //用来保存重儿子
int dep[N]; //用来保存当前节点的深度
int fa[N]; //用来保存当前节点的父亲
int tid[N]; //用来保存树中每个节点剖分后的新编号，线段树
int Rank[N];//tid反向数组，不一定需要
inline void AddEdge(int f, int t){
edges[++E] = Edge(f, t, head[f]);
head[f] = E;
}
inline void Init(int n){
tim = 0;
this -> n = n ; E = -1;
for (int i = 0; i <= n; i++) head[i] = -1;
for (int i = 0; i <= n; i++) son[i] = -1;
}
void dfs1(int u, int father, int d){
dep[u] = d;
fa[u] = father;
siz[u] = 1;
int nxt;
for(int i = head[u]; i != -1; i = nxt){
Edge &e = edges[i]; nxt = e.nxt;
if (e.to == father) continue;
dfs1(e.to, u, d + 1);
siz[u] += siz[e.to];
if(son[u]==-1 || siz[e.to] > siz[son[u]]) son[u] = e.to;
}
}
void dfs2(int u, int tp){
top[u] = tp;
tid[u] = ++tim;
Rank[tid[u]] = u;
if (son[u] == -1) return;
dfs2(son[u], tp);
int nxt;
for(int i = head[u]; i != -1; i = nxt){
Edge &e = edges[i]; nxt = e.nxt;
if(e.to == son[u] || e.to == fa[u]) continue;
dfs2(e.to, e.to);
}
}
LL query(int u, int v){
int f1 = top[u], f2 = top[v];
LL tmp = 0;
for (; f1 != f2;){
if (dep[f1] < dep[f2]){
swap(f1, f2);
swap(u, v);
}
tmp += T.query(tid[f1], tid[u]);
u = fa[f1]; f1 = top[u];
}
if (dep[u] > dep[v]) swap(u, v);
return tmp + T.query(tid[u], tid[v]);
}
} g ;
树分治
poj1741模板题
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e4 + 7;
const int INF = 0x3f3f3f3f;
int n, k, ans;
struct Edge{
int from, to, w, nxt;
Edge(){}
Edge (int f, int t, int _w, int n):from(f),to(t),w(_w),nxt(n){}
} edges[N * 2];
bool vis[N];
int head[N], E, siz[N], dep[N];
void Init(){
E = 0;
memset(head, -1, sizeof(head));
memset(vis, false, sizeof(vis));
}
void AddEdge(int u,int v,int w){
edges[E] = Edge(u, v, w, head[u]);
head[u] = E++;
}
int dfssize(int u, int pre){
siz[u] = 1;
for(int i = head[u]; i != -1; i = edges[i].nxt){
Edge &e = edges[i];
if(e.to == pre || vis[e.to])continue;
siz[u] += dfssize(e.to, u);
}
return siz[u];
}
//找重心
void dfsroot(int u, int pre, int totnum, int &minn, int &root){
int maxx = totnum - siz[u];
for (int i = head[u]; i != -1;i = edges[i].nxt){
Edge &e = edges[i];
if(e.to == pre || vis[e.to]) continue;
dfsroot(e.to, u, totnum, minn, root);
maxx = max(maxx, siz[e.to]);
}
if(maxx < minn){minn = maxx; root = u;}
}
//求每个点离重心的距离
void dfsdep(int u,int pre,int dist, int &num){
dep[num++] = dist;
for(int i = head[u]; i != -1; i = edges[i].nxt){
Edge &e = edges[i];
if(e.to == pre || vis[e.to])continue;
dfsdep(e.to, u, dist + e.w, num);
}
}
//计算以u为根的子树中有多少点对的距离小于等于K
int calc(int u, int d){
//printf("calc(%d, %d)\n", u, d);
int ans = 0, num = 0;
dfsdep(u, -1, d, num);
sort(dep, dep + num);
int i = 0, j = num - 1;
for (; i < j; i++){
while (dep[i]+dep[j]>k && i<j) j--;
ans += j - i;
}
return ans;
}
void solve(int u){
int Max = N, root, minn = INF;
int totnum = dfssize(u, -1);
dfsroot(u, -1, totnum, minn, root);
ans += calc(root, 0);
vis[root] = 1;
for(int i = head[root]; i != -1; i = edges[i].nxt){
Edge &e = edges[i];
if (vis[e.to]) continue;
ans -= calc(e.to, e.w);
solve(e.to);
}
}
int main(){
///freopen("in.txt","r",stdin);
int u, v, w;
for (; ~scanf("%d%d", &n, &k) && (n|k);){
Init();
for(int i = 1; i < n; i++){
scanf("%d%d%d", &u, &v, &w);
AddEdge(u, v, w);
AddEdge(v, u, w);
}
ans = 0;
solve(1);
printf("%d\n", ans);
}
return 0;
}
图的割点、桥和双连通分支的基本概念
[点连通度与边连通度]
在一个无向连通图中，如果有一个顶点集合，删除这个顶点集合，以及这个集合中所有顶点相关联的边以
后，原图变成多个连通块，就称这个点集为 割点集合。一个图的 点连通度的定义为，最小割点集合中的顶
点数。
类似的，如果有一个边集合，删除这个边集合以后，原图变成多个连通块，就称这个点集为 割边集合。一
个图的 边连通度的定义为，最小割边集合中的边数。
[双连通图、割点与桥]
如果一个无向连通图的点连通度大于 1，则称该图是 点双连通的(point biconnected)，简称 双连通或 重连通。
一个图有割点，当且仅当这个图的点连通度为 1，则割点集合的唯一元素被称为 割点(cut point)，又叫 关节
点(articulation point)。
如果一个无向连通图的边连通度大于 1，则称该图是 边双连通的(edge biconnected)，简称双连通或重连通。
一个图有桥，当且仅当这个图的边连通度为 1，则割边集合的唯一元素被称为 桥(bridge)，又叫 关节边
(articulation edge) 。
可以看出，点双连通与边双连通都可以简称为双连通，它们之间是有着某种联系的，下文中提到的双连通，
均既可指点双连通，又可指边双连通。
[双连通分支]
在图 G 的所有子图 G’中，如果 G’是双连通的，则称 G’为 双连通子图。如果一个双连通子图 G’它不是任何一
个双连通子图的真子集，则 G’为 极大双连 通子图。 双连通分支(biconnected component)，或 重连通分支，
就是图的极大双连通子图。特殊的，点双连通分支又叫做 块。
[求割点与桥]
该算法是 R.Tarjan 发明的。对图深度优先搜索，定义DFS(u)为u 在搜索树（以下简称为树）中被遍历到的次
序号。定义 Low(u)为u 或u 的子树中能通过非父子边追溯到的最早的节点，即 DFS 序号最小的节点。根据
定义，则有：
Low(u)=Min { DFS(u) DFS(v) (u,v)为后向边(返祖边) 等价于 DFS(v)
动态规划
各种背包
背包
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100007;
struct node {
int v,w,n;
node(){}
node(int x,int y,int z){
v=x,w=y,n=z;
}
}a[N];
int f[N];
int main(){
//freopen("fuck.in","r",stdin);
int cash,n,x,y;
for (;~scanf("%d%d",&cash,&n);){
int A = 0;
for(int i=1;i<=n;i++){
scanf("%d%d",&x,&y);
for (int t=0;(1<<t)<x;t++){
int tt=1<<t;
a[++A]=node(y*tt,y*tt,1);
x -= tt;
}
if (x)a[++A]=node(y*x,y*x,1);
}
memset(f,0,sizeof(f));//01背包
for (int i=1;i<=A;i++)
for (int j=cash;j>=a[i].v;j--)
f[j]=max(f[j],f[j-a[i].v]+a[i].w);
int ans = 0;//get ans
for (int i=0;i<=cash;i++) ans=max(ans,f[i]);
printf("%d\n",ans);
}
return 0;
}
多重背包通用模板（单调队列）
int f[N];
int va[N], vb[N];//MAX_V
void pack(int V,int v,int w,int n){
if (n==0||v==0) return;
if (n==1){//01背包
for (int i=V;i>=v;--i)
f[i]=max(f[i],f[i-v]+w);
return;
}
if (n*v>=V-v+1){//多重背包(n >= V / v)
for (int i=v;i<=V;++i)
f[i]=max(f[i],f[i-v]+w);
return;
}
for (int j = 0 ; j < v ; ++j ){
int *pb = va, *pe = va - 1;
int *qb = vb, *qe = vb - 1;
for (int k=j,i=0;k<=V;k+=v,++i){
if (pe==pb+n){
if(*pb == *qb) ++qb;
++pb;
}
int tt = f[k] - i * w;
*++pe = tt;
while (qe>=qb&& *qe<tt)--qe;
*++qe = tt;
f[k] = *qb + i * w;
}
}
}
//主程序调用
memset(f,0,sizeof(f)); //pack
for (int i=1;i<=n;i++)
pack(cash,a[i].v,a[i].w,a[i].n);
int ans = 0; //getAns
for (int i=0;i<=cash;i++) ans=max(ans,f[i]);
printf("%d\n",ans);
小价值大重量背包问题
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
const int MAXN = 1005, MAXV = 1000005;
int f[MAXV];
int v[MAXN];
int w[MAXN];
const int INF = 0x3f3f3f3f;
int main(){
int T;
cin >> T;
while(T--) {
int n, V;
cin >> n >> V;
fill(f, f + MAXV, INF);
f[0] = 0;
for(int i = 0; i < n; i++) scanf("%d", v + i);
for(int i = 0; i < n; i++) scanf("%d", w + i);
for(int i = 0; i < n; i++) {
for(int j = MAXV - 1; j >= v[i] ; j--) {
f[j] = min(f[j], f[j - v[i]] + w[i]);
}
}
int ans = -1;
for(int i = MAXV - 1; i >= 0; i--) {
if (f[i] <= V) {
ans = i;
break;
}
}
if (ans != -1) printf("%d\n", ans);
else printf("No solution\n");
}
return 0;
}
输出LCS序列
#include <string.h>
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string>
#include <vector>
using namespace std;
string s1,s2;
string lcs2(string s1,string s2){
if(s1==""||s2=="")return "";
int m=s1.size() + 1;
int n=s2.size() + 1;
printf("%d %d\n",m,n);
int lcs[m][n];
memset(lcs,0,sizeof(lcs));
for(int i=1;i<m;i++)
for(int j=1;j<n;j++){
if(s1[i-1]==s2[j-1])
lcs[i][j]=lcs[i-1][j-1]+1;
else
lcs[i][j]=lcs[i-1][j]>=lcs[i][j-1]?lcs[i-1][j]:lcs[i][j-1];//取上侧或左侧的最大值
}
int i=m-2;
int j=n-2;
string ss="";
while(i!=-1&&j!=-1) {
if(s1[i]==s2[j]) {
//printf("%c\n",s1[i]);
ss+=s1[i];
i--; j--;
} else {
if(lcs[i+1][j+1]==lcs[i][j]) {
i--; j--;
} else {
if(lcs[i][j+1]>=lcs[i+1][j]) i--;
else j--;
}
}
}
reverse(ss.begin(),ss.end());//将字符串倒置
return ss;
}
int main(){
while(cin>>s1>>s2){
string s=lcs2(s1,s2);
cout << s<<endl;
}
return 0;
}
TSP旅行商问题.(状压DP)
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 16;
const int INF = 0x3f3f3f3f;
int n;
int dp[1 << MAXN][MAXN];
int d[MAXN][MAXN];
int rec(int s, int v) {
//记忆化
if (dp[s][v] >= 0) {return dp[s][v];}
//已经访问过所有节点并返回零号节点
if (s == (1 << n) - 1 && v == 0) {
return dp[s][v] = 0;
}
int res = INF;
for (int u = 0; u < n; u++) {
if (!(s >> u & 1)) {
//下一步移动到顶点U
res = min(res, rec(s | 1 << u, u) + d[v][u]);
}
}
return dp[s][v] = res;
}
int solve() {
memset(dp, -1, sizeof(dp));
printf("%d\n", rec(0, 0));
}
void floyd() {
for(int k = 0; k < n; k++) {
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if (i == j) d[i][j] = 0;
else d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
}
}
}
int main(){
int T;
cin >> T;
while(T--) {
memset(d, INF, sizeof(d));
int m;
cin >> n >> m;
for(int i = 0; i < m; i++) {
int x, y, v;
scanf("%d%d%d", &x, &y, &v);
if (d[x - 1][y - 1] > v) d[x - 1][y - 1] = d[y - 1][x - 1] = v;
}
//每个点只能走一次就是原本的TSP问题，可以走多次先求各点最短路转化为TSP
//道理比如1 - 3, 1 - 2 那么 1 要走3次， TSP求不出 但是求一波Floyd以后， 3 - 2有条路了，相当于改变了图
floyd();
solve();
}
return 0;
}
DAG序列反向DP
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 1e5+7;
const int mod = 1e9+7;
vector<int>e[maxn];
LL a[maxn],b[maxn],d[maxn];
LL ans[maxn];
int n,m;
int main(){
while(scanf("%d%d",&n,&m)!=EOF) {
for(int i = 0;i<=n;i++)
e[i].clear();
memset(d,0,sizeof(d));
memset(ans,0,sizeof(ans));
for(int i = 1;i<=n;i++)
scanf("%lld%lld",&a[i],&b[i]);
for(int i = 1;i<=m;i++){
int u,v;
scanf("%d%d",&u,&v);
e[v].push_back(u);
d[u]++;
}
queue<int>q;
for(int i = 1;i<=n;i++)
if(d[i]==0)q.push(i);
while(!q.empty()){
int u = q.front();q.pop();
for(int i = 0;i<e[u].size();i++){
int v = e[u][i];
ans[v]=(ans[v]+(ans[u]+b[u])%mod)%mod;
if(--d[v]==0)
q.push(v);
}
}
LL res = 0;
for(int i = 1;i<=n;i++)
res = (res + 1LL*ans[i]*a[i]%mod)%mod;
printf("%lld\n",res);
}
}
数位dp
/*****************
不要62
******************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef int LL;
LL a[30];
LL dp[30][2];
LL z[30] = {1};
LL n;
LL dfs(LL pos, LL stat, bool limit) {
if(pos == -1) return 1;
if(!limit && dp[pos][stat] != -1) return dp[pos][stat];
LL up = limit ? a[pos] : 9;
LL ans = 0;
for(LL i = 0; i <= up; i++) {
if(i == 2 && stat) continue;
if(i == 4) continue;
ans += dfs(pos - 1, i == 6, limit && i == a[pos]);
}
if(!limit) dp[pos][stat] = ans;
return ans;
}
LL solve(LL x) {
LL pos = 0;
while(x) {
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos - 1, 0, true);
}
int main()
{
memset(dp, -1, sizeof(dp));
LL n1, n2;
while(scanf("%d%d", &n1, &n2), n1 + n2) {
printf("%d\n", solve(n2) - solve(n1 - 1));
}
return 0;
}
/*****************
只要49
******************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
LL a[100];
LL dp[100][2];
LL z[100] = {1};
LL n;
LL dfs(LL pos, LL stat, bool limit) {
if(pos == -1) return 0;
if(!limit && dp[pos][stat] != -1) return dp[pos][stat];
LL up = limit ? a[pos] : 9;
LL ans = 0;
for(LL i = 0; i <= up; i++) {
if(stat && i == 9) {
ans += limit ? (n % z[pos] + 1) : z[pos];
} else {
ans += dfs(pos - 1, i == 4, limit && i == a[pos]);
}
}
if(!limit) dp[pos][stat] = ans;
return ans;
}
LL solve(LL x) {
LL pos = 0;
while(x) {
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos - 1, 0, true);
}
int main()
{
for(int i = 1;i < 30; i++) {
z[i] = z[i - 1] * 10;
}
memset(dp, -1, sizeof(dp));
int t;
cin >> t;
while(t--) {
cin >> n;
cout << solve(n) << endl;
}
return 0;
}
第二种方法
LL n, dp[25][3];
//dp[i][j]:长度为i，状态为j
int digit[25];
//nstatus: 0：不含49, 1：不含49但末尾是4, 2 :含49
LL DFS(int pos, int status, int limit)
{
if(pos <= 0) // 如果到了已经枚举了最后一位，并且在枚举的过程中有49序列出现
return status==2;//注意是 ==
if(!limit && dp[pos][status]!=-1) //对于有限制的询问我们是不能够记忆化的
return dp[pos][status];
LL ans = 0;
int End = limit?digit[pos]:9; // 确定这一位的上限是多少
for(int i = 0; i <= End; i++) // 每一位有这么多的选择
{
int nstatus = status; // 有点else s = statu 的意思
if(status==0 && i==4)//高位不含49，并且末尾不是4 ，现在末尾添4返回1状态
nstatus = 1;
else if(status==1 && i!=4 && i!=9)//高位不含49，且末尾是4，现在末尾添加的不是4返回0状态
nstatus = 0;
else if(status==1 && i==9)//高位不含49，且末尾是4，现在末尾添加9返回2状态
nstatus = 2;
ans+=DFS(pos-1, nstatus, limit && i==End);
}
if(!limit)
dp[pos][status]=ans;
return ans;
}
第三种方法·没有DFS的纯DP
LL dp[27][3];
int c[27];
//dp[i][j]:长度为i的数的第j种状态
//dp[i][0]:长度为i但是不包含49的方案数
//dp[i][1]:长度为i且不含49但是以9开头的数字的方案数
//dp[i][2]:长度为i且包含49的方案数
void init()
{
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for(int i = 1; i <= 20; i++)
{
dp[i][0] = dp[i-1][0]*10-dp[i-1][1];
dp[i][1] = dp[i-1][0]*1;
dp[i][2] = dp[i-1][2]*10+dp[i-1][1];
}
}
/*****************
被13整除且包含“13”
******************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
typedef int LL;
LL a[30];
LL dp[30][15][3];
LL z[30] = {1};
LL n;
LL dfs(LL pos, LL mod, LL stat, bool limit) {
if(pos == -1) return mod == 0 && stat == 2;
if(!limit && dp[pos][mod][stat] != -1) return dp[pos][mod][stat];
LL up = limit ? a[pos] : 9;
LL ans = 0;
for(LL i = 0; i <= up; i++) {
LL ns = stat;
if(stat == 0 && i == 1) ns = 1;
if(stat == 1 && i != 1) ns = 0;
if(stat == 1 && i == 3) ns = 2;
ans += dfs(pos - 1, (mod * 10 + i) % 13, ns, limit && i == a[pos]);
}
if(!limit) dp[pos][mod][stat] = ans;
return ans;
}
LL solve(LL x) {
LL pos = 0;
while(x) {
a[pos++] = x % 10;
x /= 10;
}
return dfs(pos - 1, 0, 0, true);
}
int main()
{
memset(dp, -1, sizeof(dp));
for(LL i = 1; i < 30; i++) {
z[i] = z[i - 1] * 10;
}
while(cin >> n) {
cout << solve(n)<< endl;
}
return 0;
}
计算几何(见红书)
其他
C++高精度
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int MAXN = 600;
char s1[MAXN], s2[MAXN];
struct bign
{
int len, s[MAXN];
bign (){
memset(s, 0, sizeof(s));
len = 1;
}
bign (int num) { *this = num; }
bign (const char *num) { *this = num; }
bign operator = (const int num){
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator = (const char *num){
for(int i = 0; num[i] == '0'; num++) ; //ȥǰµ¼0
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
return *this;
}
bign operator + (const bign &b) const //+
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++){
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
bign operator += (const bign &b){
*this = *this + b;
return *this;
}
void clean(){
while(len > 1 && !s[len-1]) len--;
}
bign operator * (const bign &b) //*
{
bign c;
c.len = len + b.len;
for(int i = 0; i < len; i++){
for(int j = 0; j < b.len; j++){
c.s[i+j] += s[i] * b.s[j];
}
}
for(int i = 0; i < c.len; i++){
c.s[i+1] += c.s[i]/10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator *= (const bign &b){
*this = *this * b;
return *this;
}
bign operator - (const bign &b){
bign c;
c.len = 0;
for(int i = 0, g = 0; i < len; i++){
int x = s[i] - g;
if(i < b.len) x -= b.s[i];
if(x >= 0) g = 0;
else{
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}
bign operator -= (const bign &b){
*this = *this - b;
return *this;
}
bign operator / (const bign &b){
bign c, f = 0;
for(int i = len-1; i >= 0; i--){
f = f*10;
f.s[0] = s[i];
while(f > b || f == b)
{
f -= b;
c.s[i]++;
}
}
c.len = len;
c.clean();
return c;
}
bign operator /= (const bign &b){
*this = *this / b;
return *this;
}
bign operator % (const bign &b){
bign r = *this / b;
r = *this - r*b;
return r;
}
bign operator %= (const bign &b){
*this = *this % b;
return *this;
}
bool operator < (const bign &b){
if(len != b.len) return len < b.len;
for(int i = len-1; i >= 0; i--){
if(s[i] != b.s[i]) return s[i] < b.s[i];
}
return false;
}
bool operator > (const bign &b){
if(len != b.len) return len > b.len;
for(int i = len-1; i >= 0; i--)
{
if(s[i] != b.s[i]) return s[i] > b.s[i];
}
return false;
}
bool operator == (const bign &b){
return !(*this > b) && !(*this < b);
}
string str() const{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
return res;
}
};
int main(){
bign a, b, c;
while(scanf("%s %s", s1, s2) != EOF)
{
a = bign(s1);
b = bign(s2);
c = a / b;
cout << c.str() << endl;
}
return 0;
}
Java大整数
—：在java中的基本头文件（java中叫包）
import java.io.*
importjava.util.* 我们所用的输入scanner在这个包中
importjava.math.* 我们下面要用到的BigInteger就这这个包中
二：输入与输出
读入 Scanner cin=new Scanner（System.in）
While(cin.hasNext()) //相当于C语言中的！=EOF
n = cin.nextInt(); //输入一个int型整数
n = cin.nextBigInteger(); //输入一个大整数
System.out.print(n); //输出n但不换行
System.out.println(); //换行
System.out.println(n); //输出n并换行
System.out.printf(“%d\n”,n); //类似C语言中的输出
三：定义变量
定义单个变量：
int a,b,c; //和C++ 中无区别
BigInteger a; //定义大数变量a
BigIntegerb= new BigInteger("2"); //定义大数变量 b赋值为 2；
BigDecimaln； //定义大浮点数类 n；
定于数组：
int a[]= new int[10] //定义长度为10的数组a
BigInteger b[] =new BigInteger[100] //定义长度为100的数组a
四：表示范围
布尔型 boolean 1 true,false false
字节型 byte 8 -128-127 0
字符型 char 16 ‘\u000’-\uffff ‘\u0000’
短整型 short 16 -32768-32767 0
整型 int 32 -2147483648,2147483647 0
长整型 long 64 -9.22E18,9.22E18 0
浮点型 float 32 1.4E-45-3.4028E+38 0.0
双精度型 double 64 4.9E-324,1.7977E+308 0.0
BigInteger任意大的数，原则上只要你的计算机内存足够大，可以有无限位
五：常用的一些操作
A=BigInteger.ONE; //把0赋给A
B=BigInteger. valueOf (3); //把3赋给B；
A[i]=BigInteger. valueOf (10); //把10赋给A[i]
c=a.add(b) //把a与b相加并赋给c
c=a.subtract(b) //把a与b相减并赋给c
c=a.multiply(b) //把a与b相乘并赋给c
c=a.divide(b) //把a与b相除并赋给c
c=a.mod(b) // 相当于a%b
a.pow(b) //相当于a^b
a.compareTo(b)： //根据该数值是小于、等于、或大于a 返回 -1、0 或 1；
a.equals(b)： //判断两数是否相等，也可以用compareTo来代替；
a.min(b)，a.max(b)： //取两个数的较小、大者；
例题：hdu5920
题意：求一个数用50个以内的回文数加起来的方案
import java.math.*;
//import java.io.*;
import java.util.*;
public class Main {
public static BigInteger fanzhuan(BigInteger n){
//System.out.println("fanzhuan" + n);
int k = String.valueOf(n).length();
BigInteger ret = BigInteger.ZERO;
for (int i=1;i<=k;i++){
ret = ret.add(n.mod(BigInteger.TEN));
ret = ret.multiply(BigInteger.TEN);
n = n.divide(BigInteger.TEN);
}
ret = ret.divide(BigInteger.TEN);
return ret;
}
public static void main(String[] argv){
Scanner cin = new Scanner(System.in);
int T =cin.nextInt();
for (int cas=1;cas<=T;cas++){
BigInteger n = cin.nextBigInteger();
int N = String.valueOf(n).length();
int A = 0;
BigInteger ans[] = new BigInteger[55];
for (;N>1;){
//System.out.println("n=" + n);
BigInteger one0 = BigInteger.TEN.pow(N>>1);
BigInteger half = n.divide(one0);
if (N<=2){
if (N==1){
ans[++A] = n;
n = BigInteger.ZERO;
break;
}else {//2 wei
if (n.compareTo(BigInteger.valueOf(19))==0){
ans[++A] = BigInteger.valueOf(11);
ans[++A] = BigInteger.valueOf( 8);
n = BigInteger.ZERO;
break;
}else if (n.compareTo(BigInteger.valueOf(19))==-1){
ans[++A] = BigInteger.valueOf(9);
ans[++A] = n.subtract(BigInteger.valueOf(9));
n = BigInteger.ZERO;
break;
}//else continue;
}
}
half = half.subtract(BigInteger.ONE);
//System.out.println("half=" + half);
BigInteger fan = fanzhuan(half);
if (N%2>0) fan = fan.mod(one0);
//System.out.println("fan=" + fan);
BigInteger jan = half.multiply(one0).add(fan);
//System.out.println("jan=" + jan);
ans[++A] = jan ;
n = n.subtract(jan);
N = String.valueOf(n).length();
}
if (n.compareTo(BigInteger.ZERO)==1)ans[++A] = n;
System.out.printf("Case #%d:\n%d\n",cas,A);
for (int i=1;i<=A;i++){
System.out.println(ans[i]);
}
}
cin.close();
}
}
头文件
#include <iostream> #pragma comment(linker, "/STACK:1024000000,1024000000") #include <stdio.h> #include <fstream> #include <iomanip> #include <cmath> #include <string> #include <string.h> #include <sstream> #include <cctype> #include <climits> #include <set> #include <map> #include <deque> #include <queue> #include <vector> #include <iterator> #include <algorithm> #include <stack> #include <functional> /*int类型最大值INT_MAX，short最大值为SHORT_MAX long long最大值为LONG_LONG_MAX*/ //cout << "OK" << endl; #define _clr(x,y) memset(x,y,sizeof(x)) #define _inf(x) memset(x,0x3f,sizeof(x)) #define pb push_back #define mp make_pair #define FORD(i,a,b) for (int i=(a); i<=(b); i++) #define FORP(i,a,b) for (int i=(a); i>=(b); i--) #define REP(i,n) for (int i=0; i<(n); i++) using namespace std; const int INF = 0x3f3f3f3f; const double eps = 1e-8; const double EULER = 0.577215664901532860; const double PI = 3.1415926535897932384626; const double E = 2.71828182845904523536028; typedef long long LL;
输入挂
const int BUFSIZE = 100 * 1024 * 1024;
char Buf[BUFSIZE + 1], *buf = Buf;
template<class T>
void read(T &a){
for(a=0; *buf<'0' || *buf>'9'; buf++);
while(*buf>='0' && *buf<='9'){
a = a*10+(*buf-'0'); buf++;
}
}
对拍模板,freopen
FOR ACM OI
在linux的shell脚本对拍命令
执行方法：在终端下，进入当前目录，输入”sh ./nick.sh”,（其中nick.sh为当前shell脚本名）
ubuntu14.04下实测成功
while true; do ./make>tmp.in #出数据 ./tmp<tmp.in>tmp.out #被测程序 ./tmp2<tmp.in>tmp2.out #正确（暴力）程序 if diff tmp.out tmp2.out; then #比较两个输出文件 printf AC #结果相同显示AC else echo WA #结果不同显示WA，并退出 #cat tmp.out tmp2.out exit 0 fi #if的结束标志,与C语言相反，0为真 done # while的结束标志 #BY NICK WONG 2014-08-29 #在终端下，进入当前目录，输入"sh ./nick.sh",（其中nick.sh为当前shell脚本名） '#'表示单行注释 #diff在两文件相同时返回空串
freopen的关闭
freopen("in.txt", "r", stdin);
fclose(stdin);
freopen("CON", "r", stdin);
int m;
cin >> m;
cout << m;
/*
windows 下对拍
:again D:\cb-work4\gen\bin\Debug\gen.exe D:\cb-work4\duiA\bin\Debug\duiA.exe D:\cb-work4\duiB\bin\Debug\duiB.exe fc C:\Users\admin\Desktop\duipai\out1.txt C:\Users\admin\Desktop\duipai\out2.txt if not errorlevel 1 goto again pause
linux 下对拍
if diff test.out test.ans;then echo AC else echo WA exit 0 fi done */
来源：https://www.cnblogs.com/cww97/p/7533949.html