枚举+递推

由于按键的先后顺序对最后的结果并无影响，且重复按键只能做无用功，于是可以考虑对每个点按0/1次。通过枚举第一行的按键情况，可以使用二进制 $[0,2^{32})$ 表示按键情况，依次对之后的所有0进行按键反转。

#include <bits/stdc++.h>
#include <cstring>
using namespace std;
const int N =6, inf = 0x3f3f3f3f;
int t, n = 5, m, d[][2] = {0,0,-1, 0, 1, 0, 0, -1, 0, 1};
char s[N][N],g[N][N];

void turn(int x, int y) {
    for (int i = 0; i < 5; i++) {
        int a = x + d[i][0], b = y + d[i][1];
        if (a < 0 || a >= n || b < 0 || b >= n) continue;
        g[a][b] ^= 1;
    }
}
void solve1(){
    int ans = inf;
    for (int k = 0; k < 32; k++) {
        memcpy(g,s,sizeof s);
        int v = 0, tt = 0;
        for (int j = 0; j < 5; j++)
            if (k >> j & 1) turn(0, j), v++;
        for (int i = 1; i < 5; i++)
            for (int j = 0; j < 5; j++)
                if (g[i-1][j]=='0') turn(i, j), v++;
        for (int j = 0; j < n; j++) tt+=g[4][j]-'0';
        if (v<=6 && tt==5) ans = min(ans, v);
    }
    cout << (ans == inf ? -1 : ans) << endl;
}
int main() {
    cin >> t;
    while (t--) {
        for (int i = 0; i < n; i++) cin >> s[i];
        solve1();
    }
}