先枚举第一行，按第一行灯泡，保持不变，依次枚举下一行，当最后一行仍然有灯泡熄灭，则不符合题意；
若6步以内仍无法使所有灯泡变凉则输出-1
用二进制来表示1为开，0为关，改变其值只要异或1即可
用偏移量表示可能改变的灯泡
用缓冲数组存储当前灯泡的状态
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 6;
int n;
char g[N][N],backup[N][N];
int dx[5] = {-1,0,1,0,0},dy[5] = {0,1,0,-1,0};

void turn(int x,int y){
    for(int i = 0;i < 5;++i){
        int a = x + dx[i],b = y + dy[i];
        if(a < 0 || a >= 5 || b < 0 || b >= 5)  continue;
        g[a][b] ^= 1;
    }
}

int main(){
    cin >> n;
    while(n--){
        for(int i = 0;i < 5;++i)    cin >> g[i];
        int res = 10;
        for(int op = 0;op < 32;++op){
            memcpy(backup,g,sizeof g);
            int step = 0;
            for(int i = 0;i < 5;++i){
                if(op >> i & 1){
                    step++;
                    turn(0,i);
                }
            }
            for(int i = 0;i < 4;++i){
                for(int j = 0;j < 5;++j){
                    if(g[i][j] == '0'){
                        step++;
                        turn(i + 1,j);
                    }
                }
            }
            bool dark = false;
            for(int i = 0;i < 5;++i){
                if(g[4][i] == '0'){
                    dark = true;
                    break;
                }
            }
            if(!dark)   res = min(res,step);
            memcpy(g,backup,sizeof g);
        }
        if(res > 6) res = -1;
        cout << res << endl;
    }
    return 0;
}