95.费解的开关

95.费解的开关

用枚举的思想，把第一行先枚举了

（通过：

for (int op = 0; op < 32; op ++ )

for (int i = 0; i < 5; i ++ )

if (op >> i & 1)

）

根据第i行去trun第i+1行来改变第i行

trun的改变利用了偏移量来简化

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 6;
char g[N][N], backup[N][N];
int dx[] = {0, 0, 0, -1, 1}, dy[] = {0, -1, 1, 0, 0};

void getg()
{
    for (int i = 0; i < 5; i ++ ) cin >> g[i];
    return;
}
void print()
{
    for (int i = 0; i < 5; i ++ ) cout << backup[i] << endl;
    cout << endl;
    return;
}
void bkup()
{
    memcpy(backup, g, sizeof g);
    // for (int i = 0; i < 5; i ++ ) backup[i] = g[i];
    return;
}

void trun(int x, int y)
{
    int a, b;
    for (int i = 0; i < 5; i ++ )
    {
        a = x + dx[i], b = y + dy[i];
        if (a >= 0 && a < 5 && b >= 0 && b < 5) backup[a][b] ^= 1;
    }
}

int main()
{
    int T;
    cin >> T;
    while (T -- )
    {
        getg();
        int ans = 10;
        // print();
        // 遍历第一行情况
        for (int op = 0; op < 32; op ++ )
        {
            // 第一行的一种情况
            
            bkup();
            int cnt = 0;
            for (int i = 0; i < 5; i ++ )
            {
                if (op >> i & 1)
                {
                    cnt++;
                    trun(0, i);
                }
            }
            for (int i = 0; i < 4; i ++ )
            {
                for (int j = 0; j < 5; j ++ )
                {
                    if (backup[i][j] == '0')
                    {
                        cnt++;
                        trun(i + 1, j);
                    }
                }
            }
            bool drak = false;
            for (int i = 0; i < 5; i ++ ) if (backup[4][i] == '0') drak = true;
            
            if (!drak) ans = min(ans, cnt);
            
        }
        if (ans > 6) ans = -1;
        cout << ans << endl;
    }
    return 0;
}