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

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

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

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

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

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

/*
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