算法1

为了矩阵完美，需要每个四元组（上下左右对称的四个点）内的值一样
使得每个四元组一样的修改值最小，取中位数即可（绝对值不等式）

时间复杂度 $O(nm)$

参考文献

C++ 代码

略丑，加了考虑奇数的情况

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 104;
int a[N][N];
void solve()
{

    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 res = 0;
    for (int i = 1; i <= n / 2;i++)
    {
        for (int j = 1; j <= m / 2;j++)
        {
            vector<int> vec(4);
            vec[0] = a[i][j];
            vec[1] = a[n - i + 1][j];
            vec[2] = a[i][m - j + 1];
            vec[3] = a[n - i + 1][m - j + 1];
            sort(vec.begin(), vec.end());
            int avg = vec[1];
            res += abs(vec[0] - avg) + abs(vec[1] - avg) + abs(vec[2] - avg) + abs(vec[3] - avg);
        }
        if(m%2)
        {
            int b = a[i][m / 2 + 1];
            int c = a[n - i + 1][m / 2 + 1];
            res += abs(b - c);
        }
    }
    if(n%2)
    {
        for (int j = 1; j <= m/2;j++)
        {
            int b = a[n / 2 + 1][j];
            int c = a[n / 2 + 1][m - j + 1];
            res += abs(b - c);
        }
    }
    cout << res << endl;
}
int main()
{
    int t;
    cin >> t;
    while(t--)
        solve();
    return 0;
}