HDU - 2888 Check Corners(二维RMQ)

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题目大意：给你一个矩阵，Q个询问，询问的是一个矩阵内的最大值，并问这个最大值是否在询问的矩阵的四个角

解题思路：二维RMQ裸题

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 310;int n, m;int dp[N][N][9][9];int val[N][N];void init() {    for (int i = 1; i <= n; i++)        for (int j = 1; j <= m; j++) {            scanf("%d", &val[i][j]);            dp[i][j][0][0] = val[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 + (1 << i) - 1 <= n; row++)                for (int col = 1; col + (1 << j) - 1 <= m; col++) {                    //当x或y等于0的时候，就相当于一维的RMQ了                    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 if (j == 0) dp[row][col][i][j] = max(dp[row][col][i - 1][j], dp[row + (1 << (i - 1))][col][i - 1][j]);                    else dp[row][col][i][j] = max(dp[row][col][i][j - 1], dp[row][col + (1 << (j - 1))][i][j - 1]);                }        }    }}//本来一维RMQ询问的时候是一个区间，现在变成了一个矩形，所以需要四个角度int Query(int x1, int y1, int x2, int y2) {    int kx = 0, ky = 0;    while (x1 + (1 << (1 + kx)) - 1 <= x2) kx++;    while (y1 + (1 << (1 + ky)) - 1 <= y2) ky++;    int m1 = dp[x1][y1][kx][ky];    int m2 = dp[x2 - (1 << kx) + 1][y1][kx][ky];    int m3 = dp[x1][y2 - (1 << ky) + 1][kx][ky];    int m4 = dp[x2 - (1 << kx) + 1][y2 - (1 << ky) + 1][kx][ky];    return max(max(m1, m2), max(m3, m4));}void solve() {    int q;    scanf("%d", &q);    while (q--) {        int x1, y1, x2, y2;        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);        int ans = Query(x1, y1, x2, y2);        printf("%d", ans);        if (ans == val[x1][y1] || ans == val[x2][y1] || ans == val[x1][y2] || ans == val[x2][y2]) printf(" yes\n");        else printf(" no\n");    }}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        init();        solve();    }    return 0;}

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