### 题目描述

The Kingdom of JOIOI is a rectangular grid of H × W cells. In the Kingdom of JOIOI, in order to improve efficiency of administrative institutions, the country will be divided into two regions called “JOI” and “IOI.”

Since we do not want to divide it in a complicated way, the division must satisfy the following conditions:

• Each region must contain at least one cell.
• Each cell must belong to exactly one of two regions.
• For every pair of cells in the JOI region, we can travel from one to the other by passing through cells belonging to the JOI region only. When move from a cell to another cell, the two cells must share an edge.The same is true for the IOI region.
• For each row or column, if we take all the cells in that row or column, the cells belonging to each region must be connected. All the cells in a row or column may belong to the same region.

Each cell has an integer called the altitude. After we divide the country into two regions, it is expected that traveling in each region will be active. But, if the difference of the altitudes of cells are large, traveling between them becomes hard. Therefore, we would like to minimize the maximum of the difference of the altitudes between cells belonging to the same region. In other words, we would like to minimize the larger value of

• the difference between the maximum and the minimum altitudes in the JOI region, and
• the difference between the maximum and the minimum altitudes in the IOI region.

### 样例

#### 输入

4 4 1 12 6 11 11 10 2 14 10 1 9 20 4 17 19 10

#### 输出

11

#### 说明

For example, in this sample input, we divide the country into two regions as follows. Here, ‘J’ denotes the JOI
region, and ‘I’ denotes the IOI region.

J J J I
J J J I
J J I I
J I I I

The following division does not satisfy the condition because, in the third column from left, cells with ‘I’ are
not connected.

J J I I
J J J I
J J J I
J I I I

### 思路 ### 代码

#include <bits/stdc++.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define mid ((l+r)>>1)
#define maxn 2010
int n,m,a[maxn][maxn];
int mx,mn=INT_MAX;
bool check(int fit){
int lim=0;
for (int i=1;i<=n;i++){
for (int j=1;j<=m;j++)
if (a[i][j]<mx-fit)
lim=max(lim,j+1);
for (int j=1;j<lim;j++)
if (a[i][j]>mn+fit)
return false;
}
return true;
}
int solve(){
int l=0,r=mx-mn;
while(l<r){
if (check(mid)) r=mid;
else l=mid+1;
}
return l;
}
void flip_l(){
for (int j=1;j<=m;++j)
for (int i=1;i<=n/2;++i)
std::swap(a[n-i+1][j],a[i][j]);
}
void flip_p(){
for (int i=1;i<=n;++i)
for (int j=1;j<=m/2;j++)
std::swap(a[i][m-j+1],a[i][j]);
}
int main(){
scanf("%d%d",&n,&m);
for (int i=1;i<=n;++i)
for (int j=1;j<=m;++j)
scanf("%d",&a[i][j]),
mx=max(mx,a[i][j]),
mn=min(mn,a[i][j]);
int res=solve();
flip_l();
res=min(res,solve());
flip_p();
res=min(res,solve());
flip_l();
res=min(res,solve());
printf("%d\n",res);
return 0;
}