数列分块入门 1
区间加 + 单点查询
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 5e4 + 10;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
int A[N], Add[N], bel[N];
int n, block, cnt;
void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w;
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w;
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w;
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1;
if(n % block) cnt = n / block + 1;
else cnt = n / block;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << A[r] + Add[bel[r]] << endl;
}
return 0;
}数列分块入门 2
区间加法，询问区间内小于某个值 x 的元素个数
用B[]记录A[], B[]数组中为排好序的A[]的映射
那么每次可以对每一块进行二分查找
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 5e4 + 10;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
int A[N], B[N], Add[N], bel[N];
int n, block, cnt;
void Work_sort(int x) {
int l = (x - 1) * block + 1, r = min(l + block - 1, n);
for(int i = l; i <= r; i ++) B[i] = A[i];
sort(B + l, B + r + 1);
}
void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) A[i] += w;
Work_sort(bel[x]);
}
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w; Work_sort(bel[y]);
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w;
}
inline int Calc(int x, int w) {
int l = (x - 1) * block + 1, r = min(l + block - 1, n), ret = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(B[mid] + Add[x] < w) ret = mid, l = mid + 1;
else r = mid - 1;
}
return ret ? (ret - (x - 1) * block) : 0;
}
inline int Sec_A(int x, int y, int w) {
int ret(0);
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w) ret ++;
return ret;
} else {
for(int i = x; i <= bel[x] * block; i ++) if(A[i] + Add[bel[x]] < w) ret ++;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++)
if(A[i] + Add[bel[y]] < w)
ret ++;
}
for(int i = bel[x] + 1; i < bel[y]; i ++)
ret += Calc(i, w);
return ret;
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1;
if(n % block) cnt = n / block + 1;
else cnt = n / block;
for(int i = 1; i <= cnt; i ++) Work_sort(i);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r, c * c) << "\n";
}
return 0;
}数列分块入门 3
区间加法，询问区间内小于某个值 x 的前驱
与2类似，二分查找
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
const int oo = 999999999;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
int A[N], B[N] = {-1}, Add[N], bel[N];
int n, block, cnt;
void Work_sort(int x) {
int l = (x - 1) * block + 1, r = min(l + block - 1, n);
for(int i = l; i <= r; i ++) B[i] = A[i];
sort(B + l, B + r + 1);
}
void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) A[i] += w;
Work_sort(bel[x]);
}
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w; Work_sort(bel[x]);
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w; Work_sort(bel[y]);
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w;
}
inline int Calc(int x, int w) {
int l = (x - 1) * block + 1, r = min(l + block - 1, n), ret = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(B[mid] + Add[x] < w) ret = mid, l = mid + 1;
else r = mid - 1;
}
return B[ret] + Add[x];
}
inline int Sec_A(int x, int y, int w) {
int ret = -1;
if(bel[x] == bel[y]) {
for(int i = x; i <= y; i ++) if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];
return ret;
} else {
for(int i = x; i <= bel[x] * block; i ++)
if(A[i] + Add[bel[x]] < w && A[i] + Add[bel[x]] > ret) ret = A[i] + Add[bel[x]];
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++)
if(A[i] + Add[bel[y]] < w && A[i] + Add[bel[y]] > ret) ret = A[i] + Add[bel[y]];
}
for(int i = bel[x] + 1; i < bel[y]; i ++) {
int imp = Calc(i, w);
if(imp < w && imp > ret) ret = imp;
}
return ret;
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1;
if(n % block) cnt = n / block + 1;
else cnt = n / block;
for(int i = 1; i <= cnt; i ++) Work_sort(i);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r, c) << "\n";
}
return 0;
}数列分块入门 4
区间加法，区间求和
没什么好说的
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 5e4 + 10;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
#define LL long long
LL A[N], Add[N], bel[N], W[N];
int n, block, cnt, Mod;
void Sec_G(int x, int y, int w) {
if(bel[x] == bel[y]) for(int i = x; i <= y; i ++) A[i] += w, W[bel[x]] += w;
else {
for(int i = x; i <= bel[x] * block; i ++) A[i] += w, W[bel[x]] += w;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w, W[bel[y]] += w;
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w, W[i] += w * block;
}
inline int Sec_A(int x, int y) {
LL ret = 0;
if(bel[x] == bel[y])
for(int i = x; i <= y; i ++)
ret += (A[i] + Add[bel[x]]) % Mod;
else {
for(int i = x; i <= bel[x] * block; i ++) ret += (A[i] + Add[bel[x]]) % Mod;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) ret += (A[i] + Add[bel[y]]) % Mod;
}
for(int i = bel[x] + 1; i < bel[y]; i ++) ret += W[i] % Mod;
return ret% Mod;
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, W[bel[i]] += A[i];
if(n % block) cnt = n / block + 1;
else cnt = n / block;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read(); Mod = c + 1;
if(!opt) Sec_G(l, r, c);
else cout << Sec_A(l, r) << endl;
}
return 0;
}数列分块入门 5
区间开方，区间求和
一个数(合理)开几次根后就是0/1了
因此，只需记录每块的最大值，如果最大值是0/1就没必要开根
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 5e4 + 10;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
#define LL long long
LL A[N], Add[N], bel[N], W[N], Max[N];
int n, block, cnt, Mod;
inline void Sec_G(int x, int y) {
if(bel[x] == bel[y]) {
if(!Max[bel[x]]) return ;
if(Max[bel[x]] == 1) return ;
for(int i = x; i <= y; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[x]] -= C; A[i] = (int) sqrt(A[i]);
}
LL Max_A = 0;
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[x]] = Max_A;
return ;
}
else {
if(Max[bel[x]] && Max[bel[x]] != 1) {
for(int i = x; i <= bel[x] * block; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[x]] -= C;
A[i] = (int) sqrt(A[i]);
}
LL Max_A = 0;
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[x]] = Max_A;
}
if(Max[bel[y]] && Max[bel[y]] != 1) {
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) {
int C = A[i] - (int) sqrt(A[i]);
W[bel[y]] -= C;
A[i] = (int) sqrt(A[i]);
}
LL Max_A = 0;
for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) Max_A = max(Max_A, A[i]);
Max[bel[y]] = Max_A;
}
}
for(int i = bel[x] + 1; i < bel[y]; i ++) {
if(!Max[i] || Max[i] == 1) continue ;
LL Max_A = 0;
for(int j = (i - 1) * block + 1; j <= i * block; j ++) {
int C = A[j] - (int) sqrt(A[j]);
W[i] -= C;
A[j] = (int) sqrt(A[j]);
Max_A = max(Max_A, A[j]);
}
Max[i] = Max_A;
}
}
inline int Sec_A(int x, int y) {
LL ret = 0;
if(bel[x] == bel[y] && Max[bel[x]]) for(int i = x; i <= y; i ++) ret += A[i];
else {
for(int i = x; i <= bel[x] * block && Max[bel[x]]; i ++) ret += A[i];
for(int i = (bel[y] - 1) * block + 1; i <= y && Max[bel[y]]; i ++) ret += A[i];
}
for(int i = bel[x] + 1; i < bel[y]; i ++) ret += W[i];
return ret;
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, W[bel[i]] += A[i], Max[bel[i]] = max(Max[bel[i]], A[i]);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Sec_G(l, r);
else cout << Sec_A(l, r) << "\n";
}
return 0;
}数列分块入门 6
单点插入，单点询问
数据随机，分块，对于每一块开动态数组，插入 + 查询比较容易实现
如果数据不随机，就有可能加到同一块中的数较多，影响效率
这样可以进行一定的插入操作之后重新分块
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
using namespace std;
const int N = 1e5 + 10;
int A[N << 1], n;
vector <int> Vec[350];
int block, bel[N];
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
inline void Ins(int x, int a) {
int now_size(0), Whi;
for(int i = 1; ; i ++) {
int Size = Vec[i].size();
now_size += Size;
if(now_size >= x) {
Whi = i;
x -= (now_size - Size);
break;
}
}
Vec[Whi].insert(Vec[Whi].begin() + x - 1, a);
}
inline int Poi_A(int x) {
int Whi_, now_size(0);
for(int i = 1; ; i ++) {
int Size = Vec[i].size();
now_size += Size;
if(now_size >= x) {
int iii = x - (now_size - Size);
return Vec[i][iii - 1];
}
}
}
int main() {
n = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) A[i] = read();
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1;
for(int i = 1; i <= n; i ++) Vec[bel[i]].push_back(A[i]);
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(!opt) Ins(l, r);
else cout << Poi_A(r) << endl;
}
return 0;
}数列分块入门 7
区间乘法，区间加法，单点询问
先乘后加，乘的时候相应的加法标记也要乘
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
using namespace std;
const int N = 1e5 + 10;
const int Mod = 1e4 + 7;
#define LL long long
LL A[N], Mul[N], Add[N], bel[N];
int n, cnt, block;
#define gc getchar()
inline int read() {
int x = 0, f = 1; char c = gc;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = gc;}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
inline void Sec_Add(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[x]] + Add[bel[x]]);
Add[bel[x]] = 0; Mul[bel[x]] = 1;
for(int i = x; i <= y; i ++) A[i] += w, A[i] %= Mod;
return ;
} else {
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = 0; Mul[bel[x]] = 1;
for(int i = x; i <= bel[x] * block; i ++) A[i] += w, A[i] %= Mod;
for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[y]] = 0; Mul[bel[y]] = 1;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] += w, A[i] %= Mod;
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] += w, Add[i] %= Mod;
}
inline void Sec_Mul(int x, int y, int w) {
if(bel[x] == bel[y]) {
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = 0; Mul[bel[x]] = 1;
for(int i = x; i <= y; i ++) A[i] = (A[i] * w) % Mod;
return ;
} else {
for(int i = (bel[x] - 1) * block + 1; i <= bel[x] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[x]] = 0; Mul[bel[x]] = 1;
for(int i = x; i <= bel[x] * block; i ++) A[i] = (A[i] * w) % Mod;
for(int i = (bel[y] - 1) * block + 1; i <= bel[y] * block; i ++) A[i] = (A[i] * Mul[bel[i]] + Add[bel[i]]) % Mod;
Add[bel[y]] = 0; Mul[bel[y]] = 1;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] = (A[i] * w) % Mod;
}
for(int i = bel[x] + 1; i < bel[y]; i ++) Add[i] = (Add[i] * w) % Mod, Mul[i] = (Mul[i] * w) % Mod;
}
int main() {
n = read();
for(int i = 1; i <= n; i ++) A[i] = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1, Mul[i] = 1;
int T = n;
while(T --) {
int opt = read(), l = read(), r = read(), c = read();
if(opt == 0) Sec_Add(l, r, c);
else if(opt == 1) Sec_Mul(l, r, c);
else cout << (A[r] * Mul[bel[r]] + Add[bel[r]]) % Mod << "\n";
}
return 0;
}数列分块入门 8
暴力
区间修改没有什么难度，这题难在区间查询比较奇怪，因为权值种类比较多，似乎没有什么好的维护方法。
模拟一些数据可以发现，询问后一整段都会被修改，几次询问后数列可能只剩下几段不同的区间了。
我们思考这样一个暴力，还是分块，维护每个分块是否只有一种权值，区间操作的时候，对于同权值的一个块就O(1)统计答案，否则暴力统计答案，并修改标记，不完整的块也暴力。
这样看似最差情况每次都会耗费O(n)的时间，但其实可以这样分析：
假设初始序列都是同一个值，那么查询是O(√n)，如果这时进行一个区间操作，它最多破坏首尾2个块的标记，所以只能使后面的询问至多多2个块的暴力时间，所以均摊每次操作复杂度还是O(√n)。
换句话说，要想让一个操作耗费O(n)的时间，要先花费√n个操作对数列进行修改。
初始序列不同值，经过类似分析后，就可以放心的暴力啦。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
const int N = 1e5 + 10;
int bel[N], A[N], bec[N];
int n;
int block;
#define gc getchar()
inline int read() {
int x = 0; char c = gc;
while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x;
}
int Sec_A(int x, int y, int c) {
int ret(0);
if(bel[x] == bel[y]) {
if(bec[bel[x]] == c) ret = y - x + 1;
else if(bec[bel[x]] == -1) for(int i = x; i <= y; i ++) if(A[i] == c) ret ++;
if(~ bec[bel[x]]) for(int i = (bel[x] - 1) * block + 1; i < x; i ++) A[i] = bec[bel[x]];
if(~ bec[bel[x]]) for(int i = y + 1; i <= bel[x] * block; i ++) A[i] = bec[bel[x]];
for(int i = x; i <= y; i ++) A[i] = c;
bec[bel[x]] = -1;
} else {
if(bec[bel[x]] == c) ret += bel[x] * block - x + 1;
else if(bec[bel[x]] == -1) for(int i = x; i <= bel[x] * block; i ++) if(A[i] == c) ret ++;
if(bec[bel[y]] == c) ret += y - ((bel[y] - 1) * block);
else if(bec[bel[y]] == -1) for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) if(A[i] == c) ret ++;
for(int i = bel[x] + 1; i < bel[y]; i ++) {
if(bec[i] == c) ret += block;
else if(bec[i] == -1)
for(int j = (i - 1) * block + 1; j <= i * block; j ++)
if(A[j] == c) ret ++;
}
if(~ bec[bel[x]]) for(int i = (bel[x] - 1) * block + 1; i < x; i ++) A[i] = bec[bel[i]];
for(int i = x; i <= bel[x] * block; i ++) A[i] = c;
bec[bel[x]] = -1;
for(int i = (bel[y] - 1) * block + 1; i <= y; i ++) A[i] = c;
if(~ bec[bel[y]]) for(int i = y + 1; i <= bel[y] * block; i ++) A[i] = bec[bel[i]];
bec[bel[y]] = -1;
for(int i = bel[x] + 1; i < bel[y]; i ++) bec[i] = c;
}
return ret;
}
int main() {
n = read();
for(int i = 1; i <= n; i ++) bec[i] = -1;
for(int i = 1; i <= n; i ++) A[i] = read();
block = sqrt(n);
for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / block + 1;
int T = n;
while(T --) {
int l = read(), r = read(), c = read();
cout << Sec_A(l, r, c) << "\n";
}
return 0;
}数列分块入门 9
区间众数查询
陈立杰区间众数解题报告
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define mod 10007
#define pi acos(-1)
#define inf 0x7fffffff
#define ll long long
using namespace std;
ll read() {
ll x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9') {
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9') {
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
int n,blo,id;
int v[50005],bl[50005];
int f[505][505];
map<int,int>mp;
int val[50005],cnt[50005];
vector<int>ve[50005];
void pre(int x) {
memset(cnt,0,sizeof(cnt));
int mx=0,ans=0;
for(int i=(x-1)*blo+1; i<=n; i++) {
cnt[v[i]]++;
int t=bl[i];
if(cnt[v[i]]>mx||(cnt[v[i]]==mx&&val[v[i]]<val[ans]))
ans=v[i],mx=cnt[v[i]];
f[x][t]=ans;
}
}
int query(int l,int r,int x) {
int t=upper_bound(ve[x].begin(),ve[x].end(),r)-lower_bound(ve[x].begin(),ve[x].end(),l);
return t;
}
int query(int a,int b) {
int ans,mx;
ans=f[bl[a]+1][bl[b]-1];
mx=query(a,b,ans);
for(int i=a; i<=min(bl[a]*blo,b); i++) {
int t=query(a,b,v[i]);
if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;
}
if(bl[a]!=bl[b])
for(int i=(bl[b]-1)*blo+1; i<=b; i++) {
int t=query(a,b,v[i]);
if(t>mx||(t==mx&&val[v[i]]<val[ans]))ans=v[i],mx=t;
}
return ans;
}
int main() {
n=read();
blo=200;
for(int i=1; i<=n; i++) {
v[i]=read();
if(!mp[v[i]]) {
mp[v[i]]=++id;
val[id]=v[i];
}
v[i]=mp[v[i]];
ve[v[i]].push_back(i);
}
for(int i=1; i<=n; i++)bl[i]=(i-1)/blo+1;
for(int i=1; i<=bl[n]; i++)pre(i);
for(int i=1; i<=n; i++) {
int a=read(),b=read();
if(a>b)swap(a,b);
printf("%d\n",val[query(a,b)]);
}
return 0;
}分块算法小结：
暴力算法
时间复杂度可以
空间允许
优美
来源：oschina
链接：https://my.oschina.net/u/4305496/blog/3988941