hdu 2888 Check Corners(RMQ)

题目链接：hdu 2888 Check Corners

题目大意：给定一个矩阵，每次查询矩阵中的最大值，并且判断该最大值是否在所查询的角落上。

解题思路：一开始用线段树，一维RMQ都超时了，然后换成了二维的RMQ，结果MLE，dp数组换成9∗9就过了。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 305;

int N, M, Q, g[maxn][maxn], dp[maxn][maxn][9][9];

void rmq_init(int n, int m) {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++)
            dp[i][j][0][0] = g[i][j];
    }

    for (int x = 0; (1<<x) <= n; x++)
        for (int y = 0; (1<<y) <= m; y++)
            if (x + y)
                for (int i = 1; i + (1<<x) - 1 <= n; i++)
                    for (int j = 1; j + (1<<y) - 1 <= m; j++) {
                        if (x)
                            dp[i][j][x][y] = max(dp[i][j][x-1][y], dp[i+(1<<(x-1))][j][x-1][y]);
                        else
                            dp[i][j][x][y] = max(dp[i][j][x][y-1], dp[i][j+(1<<(y-1))][x][y-1]);
                    }
}

int rmq_query(int x1, int y1, int x2, int y2) {
    int x = 0, y = 0;
    while ((1<<(x+1)) <= x2 - x1 + 1) x++;
    while ((1<<(y+1)) <= y2 - y1 + 1) y++;
    x2 = x2 - (1<<x) + 1;
    y2 = y2 - (1<<y) + 1;

    return max( max(dp[x1][y1][x][y], dp[x2][y1][x][y]), max(dp[x1][y2][x][y], dp[x2][y2][x][y]));
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        for (int i = 1; i <= N; i++) {
            for (int j = 1; j <= M; j++)
                scanf("%d", &g[i][j]);
        }
        rmq_init(N, M);

        scanf("%d", &Q);
        int x1, y1, x2, y2;
        while (Q--) {
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            int ans = rmq_query(x1, y1, x2, y2);
            bool flag = false;
            if (ans == g[x1][y1] || ans == g[x1][y2] || ans == g[x2][y1] || ans == g[x2][y2])
                flag = true;
            printf("%d %s\n", ans, flag ? "yes" : "no");
        }
    }
    return 0;
}