You are given a grid consisting of n rows each of which is dived into m columns. The rows are numbered from 1 to n from top to bottom, and the columns are numbered from 1 to m from left to right. Each cell is identified by a pair (x, y), which means that the cell is located in the row x and column y.
Your goal is to delete a row x (1 < x < n) and a column y (1 < y < m), after this the matrix will be divided into 4 parts. Let us define the beauty of each part as the maximum value inside it. Your task is to choose x and y such that the difference between the largest and smallest beauties is as minimal as possible. Can you?
Input
The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.
The first line of each test case contains two integers n and m (3 ≤ n, m ≤ 500), specifying the grid size, in which n is the number of rows and m is the number of columns.
Then n lines follow, each line contains m integers, giving the grid. All values in the grid are between 1 and 109 (inclusive).
Output
For each test case, print a single line containing the minimum difference between the largest and smallest beauties after dividing the matrix into 4 parts.
Example
Input
2 3 3 9 5 8 3 4 1 6 2 7 4 4 22 7 3 11 3 1 8 9 5 2 4 3 13 5 6 9
Output
3 13
题意：选定一个元素删除其所在行所在列把图分为四部分，求一个点分割后的四部分每一部分的最大值，取出来得到四个，这四个求一下最大最小的差，让这个差尽可能的小
题解：预处理一下4个方向，枚举每个位置即可（除了边界），注意数组清0
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=510;
typedef long long ll;
const ll mod=1e9+7;
int m1[510][510],m2[510][510],m3[510][510],m4[510][510];
int n,m;
int main()
{
int T,x;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(m1,0,sizeof(m1));
memset(m2,0,sizeof(m2));
memset(m3,0,sizeof(m3));
memset(m4,0,sizeof(m4));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%d",&x);
m1[i][j]=m2[i][j]=m3[i][j]=m4[i][j]=x;
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
m1[i][j]=max(m1[i][j],max(m1[i-1][j],m1[i][j-1]));
for(int i=1;i<=n;i++)
for(int j=m;j>=1;j--)
m2[i][j]=max(m2[i][j],max(m2[i-1][j],m2[i][j+1]));
for(int i=n;i>=1;i--)
for(int j=1;j<=m;j++)
m3[i][j]=max(m3[i][j],max(m3[i+1][j],m3[i][j-1]));
for(int i=n;i>=1;i--)
for(int j=m;j>=1;j--)
m4[i][j]=max(m4[i][j],max(m4[i+1][j],m4[i][j+1]));
int ans=1e9+1;
for(int i=2;i<n;i++)
{
for(int j=2;j<m;j++)
{
int maxx=-1,minn=1e9+5;
maxx=max(maxx,m1[i-1][j-1]);
maxx=max(maxx,m2[i-1][j+1]);
maxx=max(maxx,m3[i+1][j-1]);
maxx=max(maxx,m4[i+1][j+1]);
minn=min(minn,m1[i-1][j-1]);
minn=min(minn,m2[i-1][j+1]);
minn=min(minn,m3[i+1][j-1]);
minn=min(minn,m4[i+1][j+1]);
ans=min(ans,maxx-minn);
}
}
printf("%d\n",ans);
}
return 0;
}