HDOJ 题目2888 Check Corners（二维RMQ）

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本文深入探讨了一种高效的算法用于查询矩阵中指定区域的最大值，并且特别关注了检查该区域四个角落是否包含最大值的功能。通过实例演示和代码实现，详细解释了如何利用动态规划和RMQ（Range Minimum Query）技术来解决此类问题。

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Check Corners

Time Limit: 2000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2171 Accepted Submission(s): 779

Problem Description

Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer numbers ( 1 <= i <= m, 1 <= j <= n ). Now he selects some sub-matrices, hoping to find the maximum number. Then he finds that there may be more than one maximum number, he also wants to know the number of them. But soon he find that it is too complex, so he changes his mind, he just want to know whether there is a maximum at the four corners of the sub-matrix, he calls this “Check corners”. It’s a boring job when selecting too many sub-matrices, so he asks you for help. (For the “Check corners” part: If the sub-matrix has only one row or column just check the two endpoints. If the sub-matrix has only one entry just output “yes”.)

Input

There are multiple test cases.

For each test case, the first line contains two integers m, n (1 <= m, n <= 300), which is the size of the row and column of the matrix, respectively. The next m lines with n integers each gives the elements of the matrix which fit in non-negative 32-bit integer.

The next line contains a single integer Q (1 <= Q <= 1,000,000), the number of queries. The next Q lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, print Q lines with two numbers on each line, the required maximum integer and the result of the “Check corners” using “yes” or “no”. Separate the two parts with a single space.

Sample Input

Sample Output

  
   20 no
13 no
20 yes
4 yes

Source

2009 Multi-University Training Contest 9 - Host by HIT

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一个矩阵，q次询问，从x1 y2到 x2 y2的子矩阵的最大值

写完了要提交了，系统卡了一下。。文件没了，，懒得自己再打了，就是模板，

看kuangbin大神更简单一些，收藏一下

ac代码

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

int val[310][310];
int dp[310][310][9][9];//最大值
int mm[310];//二进制位数减一，使用前初始化
void initRMQ(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 ii = 0; ii <= mm[n]; ii++)
        for(int jj = 0; jj <= mm[m]; jj++)
            if(ii+jj)
                for(int i = 1; i + (1<<ii) - 1 <= n;i++)
                    for(int j = 1; j + (1<<jj) - 1 <= m;j++)
                    {
                        if(ii)dp[i][j][ii][jj] = max(dp[i][j][ii-1][jj],dp[i+(1<<(ii-1))][j][ii-1][jj]);
                        else dp[i][j][ii][jj] = max(dp[i][j][ii][jj-1],dp[i][j+(1<<(jj-1))][ii][jj-1]);
                    }
}
//查询矩形内的最大值(x1<=x2,y1<=y2)
int rmq(int x1,int y1,int x2,int y2)
{
    int k1 = mm[x2-x1+1];
    int k2 = mm[y2-y1+1];
    x2 = x2 - (1<<k1) + 1;
    y2 = y2 - (1<<k2) + 1;
    return max(max(dp[x1][y1][k1][k2],dp[x1][y2][k1][k2]),max(dp[x2][y1][k1][k2],dp[x2][y2][k1][k2]));
}
int main()
{
    //在外面对mm数组进行初始化
    mm[0] = -1;
    for(int i = 1;i <= 305;i++)
        mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    int n,m;
    int Q;
    int r1,c1,r2,c2;
    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]);
        initRMQ(n,m);
        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
            if(r1 > r2)swap(r1,r2);
            if(c1 > c2)swap(c1,c2);
            int tmp = rmq(r1,c1,r2,c2);
            printf("%d ",tmp);
            if(tmp == val[r1][c1] || tmp == val[r1][c2] || tmp == val[r2][c1] || tmp == val[r2][c2])
                printf("yes\n");
            else printf("no\n");
        }
    }
    return 0;
}