题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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

char g[6][6];
char backup[6][6];
int dx[5] = {0, 0, 1, 0, -1}, 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)
            g[a][b] ^= 1;
    }
}
int work()
{
    int ans = 1e7;

    for (int i = 0; i < 1 << 5; i ++)//枚举第一行的操作；
    {
        memcpy(backup, g, sizeof g);
        int res = 0;

        for (int j = 0; j < 5; j ++)
            if (i >> j & 1)
            {
                turn(0, j);
                res ++;
            }

        for (int j = 0; j < 4; j ++)
            for (int k = 0; k < 5; k ++)
                if (g[j][k] == '0')
                {
                    turn(j + 1, k);
                    res ++;
                }

        bool successful = true;
        for (int j = 0; j < 5; j ++)
            if (g[4][j] == '0')
            {
                successful = false;
                break;
            }

        if (successful) ans = min(res, ans);
        memcpy(g, backup, sizeof backup);
    }
    if (ans > 6) return -1;
    return ans;
}

int main()
{
    int T;
    cin >> T;
    while(T --)
    {
        for (int i = 0; i < 5; i ++)
            for (int j = 0; j < 5; j ++)
                cin >> g[i][j];

        cout << work() << endl;
    }
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla