problem

B. Nice Matrix
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A matrix of size n×m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a1,a2,…,ak) is a palindrome, if for any integer i (1≤i≤k) the equality ai=ak−i+1 holds.

Sasha owns a matrix a of size n×m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.

Help him!

Input
The first line contains a single integer t — the number of test cases (1≤t≤10). The t tests follow.

The first line of each test contains two integers n and m (1≤n,m≤100) — the size of the matrix.

Each of the next n lines contains m integers ai,j (0≤ai,j≤109) — the elements of the matrix.

Output
For each test output the smallest number of operations required to make the matrix nice.

Example
inputCopy
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
outputCopy
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:

2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:

5 6 6 5
6 6 6 6
5 6 6 5

solution

/*
题意:
+ 给出一个n*m(100)的矩阵,每次可以修改1,求最少多少次使得矩阵行列都回文
思路:
+ 对于a[x][y],因为行列回文,所以a[x][y]=a[n-x+1][y]=a[x][m-y+1]=a[n-x+1][m-y+1],答案就是这四个的中位数。
+ 像这样找规律然后选定某种策略直接计算的题算是贪心题
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 110;
LL a[maxn][maxn];
int main(){
ios::sync_with_stdio(false);
int T; cin>>T;
while(T--){
int n, m; cin>>n>>m;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
cin>>a[i][j];

LL ans = 0;
int rl=1, rr = n;
while(rl <= rr){
int cl = 1, cr = m;
while(cl <= cr){
vector<LL>num;
num.push_back(a[rl][cl]);
if(rl!=rr)num.push_back(a[rr][cl]);
if(cl!=cr)num.push_back(a[rl][cr]);
if(cl!=cr && rl!=rr)num.push_back(a[rr][cr]);

LL tmp = 0;
sort(num.begin(),num.end());
for(int i = 0; i < num.size(); i++)
tmp += abs(num[i]-num[num.size()/2]);
ans += tmp;

cl++, cr--;
}
rl++, rr--;
}
cout<<ans<<"\n";
}
return 0;
}