HDOJ 2888Check Corners（裸二维RMQ）

Posted 2021-01-13 kannyi

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

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

题意：

给定一个n*m (1<=m,n<=300)的矩阵，每次询问左上角(r1,c1)到右下角(r2,c2)的子矩形中的最大值并输出。如果每次所询问的四个角有最大值，输出yes，否则输出no。

题解：

裸二维RMQ就直接上板子吧！

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int MAX=305;
int val[MAX][MAX];
int dp[MAX][MAX][9][9];//最大值
int mm[MAX];
void initRMQ(int n,int m)//m*n的矩阵
{
    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]);
}
int rmq(int x1,int y1,int x2,int y2)//所查询矩形区间内的最大值 左上角(x1,y1) -> 右上角(x2,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()
{
    mm[0]=-1;
    for(int i=1;i<=MAX;i++)
        mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    int n,m,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);//左上角(r1,c1) -> 右上角(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
");
            else printf("no
");
        }
    }
    return 0;
}