Check Corners

Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 858    Accepted Submission(s): 275

Problem DescriptionPaul 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”.) InputThere 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.

 OutputFor 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 Input4 4
4 4 10 7
2 13 9 11
5 7 8 20
13 20 8 2
4
1 1 4 4
1 1 3 3
1 3 3 4
1 1 1 1 Sample Output20 no
13 no
20 yes
4 yes Source2009 Multi-University Training Contest 9 – Host by HIT Recommendgaojie 题目大意：求矩阵子矩阵的最大值题解：二维RMQ代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#define maxn 309
using namespace std;int n,m,dp[maxn][maxn][][],map[maxn][maxn];
int x,y,xx,yy,q;void RMQ_pre(){
    for(int i=;i<=n;i++)
     for(int j=;j<=m;j++)
      dp[i][j][][]=map[i][j];
    int mx=log(double(n))/log(2.0);
    int my=log(double(m))/log(2.0);
    for(int i=;i<=mx;i++){
        for(int j=;j<=my;j++){
            if(i==&&j==)continue;
            for(int row=;row+(<<i)-<=n;row++){
                for(int col=;col+(<<j)-<=m;col++){
                    if(i==)
                     dp[row][col][i][j]=max(dp[row][col][i][j-],dp[row][col+(<<(j-))][i][j-]);
                    else
                     dp[row][col][i][j]=max(dp[row][col][i-][j],dp[row+(<<(i-))][col][i-][j]);
                }
            }
        }
    }
}int RMQ_2D(int x,int y,int xx,int yy){
    int kx=log(double(xx-x+))/log(2.0);
    int ky=log(double(yy-y+))/log(2.0);
    int m1=dp[x][y][kx][ky];
    int m2=dp[xx-(<<kx)+][y][kx][ky];
    int m3=dp[x][yy-(<<ky)+][kx][ky];
    int m4=dp[xx-(<<kx)+][y-(<<ky)+][kx][ky];
    return max(max(m1,m2),max(m3,m4));
}int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=;i<=n;i++)
         for(int j=;j<=m;j++)
          scanf("%d",&map[i][j]);
        RMQ_pre();
        scanf("%d",&q);
        while(q--){
            scanf("%d%d%d%d",&x,&y,&xx,&yy);
            int ans=RMQ_2D(x,y,xx,yy);
            printf("%d ",ans);
            if(ans==map[x][y]||ans==map[xx][y]||ans==map[xx][yy]||ans==map[xx][yy])
                printf("yes\n");
            else printf("no\n");
        }
    }
    return ;
}