题意：
求满足题目表述的变成传送门的最小代价
思路：
n^4暴力+剪枝，先预处理二维前缀和，之后在枚举最后一列之前的代价如果比现在所得的最小代价还要大的话就break
// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <deque>
#include <vector>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define vi vector<int>
#define endl '\n'
#define int long long
using namespace std;
const int P = 1e9 + 7;
const int maxn = 4e2 + 10;
const int maxm = 4e2 + 10;
const int INF = 0x3f3f3f3f;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
int n, m;
int sum[maxn][maxm];
int a[maxn][maxm];
void solve() {
cin >> n >> m;
memset(sum, 0, sizeof(sum));
for(int i = 1; i <= n; i++) {
string s;
cin >> s;
s = ' ' + s;
for(int j = 1; j <= m; j++) a[i][j] = s[j] - '0';
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
sum[i][j] = sum[i-1][j] + sum[i][j-1] + a[i][j] - sum[i-1][j-1];
int ans = INF;
for(int i = 1; i <= n - 4; i++) {
for(int j = 1; j <= m -3; j++) {
for(int i2 = i + 4; i2 <= n; i2++) {
int now = 0;
for(int j2 = j + 3; j2 <= m; j2++) {
int cnt;
int tmp = 0;
cnt = sum[i][j2 - 1] - sum[i][j]-sum[i-1][j2-1]+sum[i-1][j];
tmp = tmp + j2 - 1 - j - cnt;
cnt = sum[i2-1][j2-1] - sum[i2-1][j] - sum[i][j2-1]+sum[i][j];
tmp = tmp + cnt;
cnt = sum[i2][j2 - 1] - sum[i2][j]-sum[i2-1][j2-1]+sum[i2-1][j];
tmp = tmp + j2 - 1 - j - cnt;
cnt = sum[i2-1][j] - sum[i][j]-sum[i2-1][j-1]+sum[i][j-1];
tmp = tmp + i2 - 1 - i - cnt;
now = max(now, tmp);
if(now > ans) break;
cnt = sum[i2-1][j2] - sum[i][j2]-sum[i2-1][j2-1]+sum[i][j2-1];
tmp = tmp + i2 - 1 - i - cnt;
ans = min(ans, tmp);
}
}
}
}
cout << ans << endl;
}
signed main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int T; cin >> T;while(T--)
solve();
return 0;
}