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Full_of_Boys训练3总结

由 扶醉桌前 提交于 2020-05-02 10:14:53

题目来源: 2016-2017 ACM-ICPC Pacific Northwest Regional Contest

E.Enclosure

先计算出内外两个凸包，枚举大凸包上的点，在小凸包上找到两个切点。计算面积时，就相当于删掉几条原先的边，加上一个新的三角形。同时，可以注意到，如果我们按照顺时针枚举大凸包上的点，那两个切点也只可能朝顺时针方向移动，这样的话，就可以在线性时间内计算出新的切点了。以后补代码。

G.Maximum Islands

贪心方法：把原本的L四个方向的C改成W，然后剩余的C，可以运用最小割的思想，用有效点数减最小割，就是最大的答案。思想来自骑士共存。二分图用的匈牙利算法。

#include <bits/stdc++.h>
#define rg register
#define pb(x) push_back(x)
typedef long long ll;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
inline write(int x) {
    if(x>=10) write(x/10);
    putchar('0'+x%10);
}
using namespace std;
int n, m, col[45][45], dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}, vis[44][44];
char s[45][45];
vector<int> v[2];
struct node{
    int x, y;
    node(){} node(int a,int b) {x = a; y = b;}
};
inline int inb(int x, int y) {
    if(x < 1 || x > n || y < 1 || y > m) return 0;
    return 1;
}
inline void bfs1(int sx, int sy) {
    queue<node> q;
    q.push(node(sx, sy));
    col[sx][sy] = 0;
    v[0].pb((sx-1)*m+sy);
    while(!q.empty()) {
        node t = q.front(); q.pop();
        for(rg int i = 0; i < 4; ++i){
            int tx = t.x + dx[i], ty = t.y + dy[i];
            if(inb(tx, ty) && s[tx][ty] == 'C' && col[tx][ty] == -1) {
                col[tx][ty] = col[t.x][t.y]^1;
                v[col[tx][ty]].pb((tx-1)*m+ty);
                q.push(node(tx, ty));
            }
        }
    }
}
struct edge{int e, nxt;}E[55*55*4];
int h[55*55], cc;
inline void add(int u, int v) {
    E[cc].e = v; E[cc]. nxt = h[u]; h[u] = cc; ++cc;
}
inline void build() {
    memset(h, -1, sizeof(h)); cc = 0;
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j <= m; ++j) if(col[i][j] == 0){
            for(rg int k = 0; k < 4; ++k) {
                int tx = i + dx[k], ty = j + dy[k];
                if(inb(tx, ty) && col[tx][ty] == 1) add((i-1)*m+j, (tx-1)*m+ty);
            }
        }
}
inline int bfs2(int sx, int sy, char c) {
    int ans = 1;
    vis[sx][sy] = 1;
    queue<node> q;
    q.push(node(sx, sy));
    while(!q.empty()) {
        node t = q.front(); q.pop();
        for(rg int i = 0; i < 4; ++i) {
            int tx = t.x + dx[i], ty = t.y + dy[i];
            if(inb(tx, ty) && s[tx][ty] == c && !vis[tx][ty]) vis[tx][ty]=1,++ans,q.push(node(tx,ty));
        }
    }
    return ans;
}
int used[55*55], lk[55*55];
inline int dfs(int u) {
    for(rg int i = h[u]; ~i; i = E[i].nxt) if(!used[E[i].e]) {
        used[E[i].e] = 1;
        if(lk[E[i].e] == -1 || dfs(lk[E[i].e])) {
            lk[E[i].e] = u;
            return 1;
        }
    }
    return 0;
}
inline int hungray() {
    int ans = 0;
    memset(lk, -1, sizeof(lk));
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j <= m; ++j) if(col[i][j] == 0){
        memset(used, 0, sizeof(used));
        if(dfs((i-1)*m+j)) ++ans;
    }
    return ans;
}
inline int solve() {
    int ans = 0, t = 0;
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j<= m; ++j) if(s[i][j] == 'L' && !vis[i][j]) bfs2(i, j, 'L'), ++ans;
    memset(vis, 0 , sizeof(vis));
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j<= m; ++j) if(s[i][j] == 'C' && !vis[i][j]) t += bfs2(i, j, 'C');
    return ans + t - hungray();
}
int main() {
    n=read(),m=read();
    memset(col, -1, sizeof(col));
    for(rg int i = 1; i <= n; ++i) scanf(" %s",s[i]+1);
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j <= m; ++j) if(s[i][j]=='L') {
            for(rg int k = 0; k < 4; ++k) if(inb(i+dx[k],j+dy[k])&&s[i+dx[k]][j+dy[k]]=='C') {
                s[i+dx[k]][j+dy[k]] = 'W';
            }
        }
    for(rg int i = 1; i <= n; ++i)
        for(rg int j = 1; j <= m; ++j) if(s[i][j] == 'C'&&col[i][j] == -1) {
            bfs1(i, j);
        }
    build();
    write(solve());
    return 0;
}

J.Shopping

每次找到区间内最左边的小于x的数，然后%一下它，重复以上操作就行了。所以只需要实现一个区间询问最左边的小于x的值就可以了。可以证明每次操作最多log次

解法1：分块。块外暴力，块内提前排好序二分。写挫了莫名t。

解法2：线段树。维护一下区间最小值，显然如果左边的最小值小于等于x那就朝左边递归，否则右边，便可以完成这个操作。

解法3：st表。又不用修改。st表干掉一个log

code：

解法1：挫了不贴了。

解法2：

#include <cstdio>
#include <algorithm>
#define pb(x) push_back(x)
#define rg register
const int maxn = 2e5 + 100;
typedef unsigned long long ll;
inline int readint(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
inline ll readll(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
inline void write(ll x){
    if(x>=10LL) write(x/10LL);
    putchar('0'+x%10LL);
}
using namespace std;
int n, q;
ll a[maxn];
struct seg{int l, r; ll x;} tr[maxn << 2];
inline void build(int p, int l, int r) {
    tr[p].l = l; tr[p].r = r;
    if(l == r){
        tr[p].x = a[l]; return;
    }
    int mid = (l + r) >> 1;
    build(p<<1, l, mid);
    build(p<<1|1, mid+1, r);
    tr[p].x = min(tr[p<<1].x, tr[p<<1|1].x);
}
inline ll ask_mn(int p, int l, int r) {
    if(tr[p].l == l && tr[p].r == r) return tr[p].x;
    int mid = (tr[p].l + tr[p].r) >> 1;
    if(r <= mid) return ask_mn(p<<1, l, r);
    else if(l > mid) return ask_mn(p<<1|1, l, r);
    else return min(ask_mn(p<<1, l, mid), ask_mn(p<<1|1, mid+1, r));
}
inline ll fd(ll x, int L, int R) {
    if(L == R) return L;
    int mid =(L + R) >> 1;
    if(ask_mn(1, L, mid) <= x) return fd(x, L, mid);
    else return fd(x, mid+1, R);
}
inline ll solve(ll x, int L, int R){
    while(ask_mn(1, L, R) <= x) x %= a[fd(x, L, R)];
    return x;
}
int main() {
    n = readint(), q = readint();
    for(int i = 1; i <= n; ++i) a[i] = readll();
    build(1, 1, n);
    while(q--) { ll x; int L, R;
        x = readll(), L = readint(), R = readint();
        write(solve(x, L, R)); puts("");
    }
    return 0;
}

解法3：

#include <cstdio>
#include <algorithm>
#include <iostream>
#define pb(x) push_back(x)
#define rg register
const int maxn = 2e5 + 100;
typedef unsigned long long ll;
inline int readint(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
inline ll readll(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
inline void write(ll x){
    if(x>=10LL) write(x/10LL);
    putchar('0'+x%10LL);
}
using namespace std;
int n, q;
ll a[maxn], st[maxn][25];
inline void RMQ_init(){
    for(int i = 0; i <= n; ++i) st[i][0] = a[i];
    for(int j = 1; (1<<j) <= n; ++j){
        for(int i = 1; i+(1<<j)-1 <= n; ++i){
            st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]);
        }
    }
}
inline ll RMQ(int u, int v){
    int k = (int)(log(v-u+1.0)/log(2.0));
    return min(st[u][k], st[v-(1<<k)+1][k]);
}
inline ll fd(ll x, int L, int R) {
    if(L == R) return L;
    int mid =(L + R) >> 1;
    if(RMQ(L, mid) <= x) return fd(x, L, mid);
    else return fd(x, mid+1, R);
}
inline ll solve(ll x, int L, int R){
    while(RMQ(L, R) <= x) x %= a[fd(x, L, R)];
    return x;
}
int main() {
    n = readint(), q = readint();
    for(int i = 1; i <= n; ++i) a[i] = readll();
    RMQ_init();
    while(q--) { ll x; int L, R;
        x = readll(), L = readint(), R = readint();
        write(solve(x, L, R)); puts("");
    }
    return 0;
}

　　

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