HDU 2888 Check Corners 二维RMQ模板

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=2888

题意：

给一个n*m的矩阵，对于每次询问，给出左上角和右下角坐标，求这个子矩阵内的最大值，并判断最大值是否在子矩阵的四角上，是则输出yes，否则no

思路：

二维RMQ裸题，在一维RMQ上进行扩展，先求出所有一维RMQ，再去求二维，算法过程很清晰。另外这个题空间卡的挺紧的。。。

#include <bits/stdc++.h>using namespace std;const int N = 300 + 5, INF = 0x3f3f3f3f;int a[N][N], dp[N][N][9][9];//dp[row][col][i][j]:以(row,col)为起点(左上角)，向下扩展2^i，向右扩展2^j长度的矩形内的最值void ST(int n, int m){    for(int i = 1; i <= n; i++)        for(int j = 1; j <= m; j++)            dp[i][j][0][0] = a[i][j];    for(int i = 0; (1<<i) <= n; i++)        for(int j = 0; (1<<j) <= m; j++)        {            if(i == 0 && j == 0) continue;            for(int row = 1; row <= n - (1<<i) + 1; row++)                for(int col = 1; col <= m - (1<<j) + 1; col++)                    if(i == 0) dp[row][col][i][j] = max(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1]);                    else dp[row][col][i][j] = max(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j]);        }}int RMQ(int x1, int y1, int x2, int y2){    int kx = log(x2 - x1 + 1.0) / log(2.0);    int ky = log(y2 - y1 + 1.0) / log(2.0);    int t1 = dp[x1][y1][kx][ky];    int t2 = dp[x2-(1<<kx)+1][y1][kx][ky];    int t3 = dp[x1][y2-(1<<ky)+1][kx][ky];    int t4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];    return max(max(t1, t2), max(t3, t4));}int main(){    int n, m;    while(~ scanf("%d%d", &n, &m))    {        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)                scanf("%d", &a[i][j]);        ST(n, m);        int q, x1, y1, x2, y2;        scanf("%d", &q);        while(q--)        {            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            int ans = RMQ(x1, y1, x2, y2);            bool flag = false;            if(a[x1][y1] == ans || a[x1][y2] == ans || a[x2][y1] == ans || a[x2][y2] == ans) flag = true;            printf("%d %s\n", ans, flag ? "yes" : "no");        }    }    return 0;}