Codeforces Round #371 (Div. 1) D - Animals and Puzzle 二维ST表 + 二分

D - Animals and Puzzle

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define PLI pair<LL, int>
#define ull unsigned long long
using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

int a[N][N], b[N][N], Log[N], n, m;

struct ST2 {
    int dp[N][N][10][10], ty;
    void build(int n, int m, int b[N][N], int _ty) {
        ty = _ty;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                dp[i][j][0][0] = ty * b[i][j];
        for(int u = 0; u <= Log[n]; u++) {
            for(int v = 0; v <= Log[m]; v++) {
                if(!u && !v) continue;
                for(int i = 1; i+(1<<u)-1 <= n; i++) {
                    for(int j = 1; j+(1<<v)-1 <= m; j++) {
                        if(u) dp[i][j][u][v] = max(dp[i][j][u-1][v], dp[i+(1<<(u-1))][j][u-1][v]);
                        else  dp[i][j][u][v] = max(dp[i][j][u][v-1], dp[i][j+(1<<(v-1))][u][v-1]);
                    }
                }
            }
        }
    }
    int query(int x1, int y1, int x2, int y2) {
        int k1 = Log[x2-x1+1], k2 = Log[y2-y1+1];
        x2 = x2-(1<<k1)+1;
        y2 = y2-(1<<k2)+1;
        return max(max(dp[x1][y1][k1][k2], dp[x2][y1][k1][k2]),
                   max(dp[x1][y2][k1][k2], dp[x2][y2][k1][k2]));
    }
} rmq;

int main() {
    for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            scanf("%d", &a[i][j]);

    for(int i = n; i >= 1; i--) {
        for(int j = m; j >= 1; j--) {
            if(!a[i][j]) continue;
            b[i][j] = min(b[i+1][j+1], min(b[i][j+1], b[i+1][j])) + 1;
        }
    }

    rmq.build(n, m, b, 1);
    int q; scanf("%d", &q);
    while(q--) {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        int l = 1, r = min(x2-x1, y2-y1) + 1, ans = 0;
        while(l <= r) {
            int mid = l + r >> 1;
            int x3 = x2 - mid + 1, y3 = y2 - mid + 1;
            if(rmq.query(x1, y1, x3, y3) >= mid) l = mid + 1, ans = mid;
            else r = mid - 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/9727247.html