HDU 2888 Check Corners (模板题)【二维RMQ】

<题目链接>

<转载于 >>> >

题目大意：

给出一个N*M的矩阵，并且给出该矩阵上每个点对应的值，再进行Q次询问，每次询问给出代询问子矩阵的左上顶点和右下顶点，问该子矩阵的最大值是多少，并且判断该最值是否在该子矩阵的四个顶角上。

解题分析：

很明显求二维区间内的最值，需要用到二维RMQ，其中dp[i][j][k][l]表示左上角为(i,j),右下角为(i + 2 ^ k - 1, j + 2 ^ l - 1)这个矩形内的最值。注意这个四维数组不要开得太大，否则容易MLE。

#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const int maxn = 305;
 
int n, m, q, val[maxn][maxn], dp[maxn][maxn][9][9];   //dp[i][j][k][l]表示左上角为(i,j),右下角为(i + 2 ^ k - 1, j + 2 ^ l - 1)这个矩形内的最值
 
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] = val[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)   //在y轴方向比较
                            dp[i][j][x][y] = max(dp[i][j][x-1][y], dp[i+(1<<(x-1))][j][x-1][y]);
                        else   //在x轴方向比较
                            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", &val[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 == val[x1][y1] || ans == val[x1][y2] || ans == val[x2][y1] || ans == val[x2][y2])   //检查最值是否在子矩阵的四个顶角上
                flag = true;
            printf("%d %s\n", ans, flag ? "yes" : "no");
        }
    }
    return 0;
}

 

 

2018-10-20