HDU 2888 Check Corners(简单二维RMQ)

题意：给定一个n * m的矩阵，再给定q个询问，每次询问(r1，c1)为左上角，(r2,c2)为右下角的子矩形的最大值，并且判断该最大值是否出现在了这个子矩阵的4个顶角上？

思路：二维RMQ，不过要注意是无符号整数

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN =305;
unsigned int val[MAXN][MAXN];
unsigned int dmax[MAXN][MAXN][9][9];
void initRMQ(int n,int m)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            dmax[i][j][0][0] = val[i][j];
    for(int ii=0;(1<<ii)<=n;ii++)
    for(int jj=0;(1<<jj)<=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)
            dmax[i][j][ii][jj]=max(dmax[i][j][ii-1][jj] , dmax[i+(1<<(ii-1))][j][ii-1][jj]);
        else
            dmax[i][j][ii][jj]=max(dmax[i][j][ii][jj-1] , dmax[i][j+(1<<(jj-1))][ii][jj-1]);
}
unsigned int getMax(int x1,int y1,int x2,int y2)
{
    int k1=0;
    while((1<<(k1+1))<=x2-x1+1)k1++;
    int k2=0;
    while((1<<(k2+1))<=y2-y1+1)k2++;
    x2 = x2-(1<<k1)+1;
    y2 = y2-(1<<k2)+1;
    return max(max(dmax[x1][y1][k1][k2] ,dmax[x1][y2][k1][k2] ) , max(dmax[x2][y1][k1][k2] , dmax[x2][y2][k1][k2]));
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2&&n&&m)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%ud",&val[i][j]);
        initRMQ(n,m);
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int x1,y1,x2,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            unsigned int max_num = getMax(x1,y1,x2,y2);
            bool ok=false;
            if(val[x1][y1]==max_num || val[x1][y2]==max_num || val[x2][y1]==max_num ||val[x2][y2]==max_num)
                ok=true;
            printf("%d %s\n",max_num,ok?"yes":"no");
        }
    }
    return 0;
}

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

4 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 Output

20 no
13 no
20 yes
4 yes