膜法记录 二进制状态压缩

添加链接描述
思路
如用1101表示选择1，3，4行

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 25;
const ll inf = 1e9;
string s[N];
int n, m, a, b, flag, t;
int main()
{
    cin >> t;
    while(t--)
    {
        flag = 0;
        cin >> n >> m >> a >> b;
        for(int i = 0; i < n; i++)
            cin >> s[i];
        for(int i = 0; i < (1 << n); i++)
        {
            int cnt = 0;
            for(int j = 0; j < n; j++)
                if((i >> j) & 1)
                    cnt++;
            if(cnt != a)
                continue;
            cnt = 0;
            for(int j = 0; j < m; j++)
                for(int k = 0; k < n; k++)
                    if( !((i >> k) & 1) && s[k][j] == '*')
                    {
                        cnt++;
                        break;
                    }
            if(cnt <= b)
            {
                flag = 1;
                break;
            }
        }
        if(flag)
            cout << "yes" << endl;
        else
            cout << "no" << endl;
    }
    return 0;
}