hdu 1760 DFS+博弈

0代表可放 1带表不能放 每次放一个2*2的方块 不能放者败
如果先手必胜则输出Yes

必胜态：从当前状态所能到达的状态中存在一个必败态
必败态：从当前状态所能达到的状态全部是必胜态

Sample Input
4 4
0000
0000
0000
0000
4 4
0000
0010
0100
0000

Sample Output
Yes
No

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <string>
# include <cmath>
# include <queue>
# include <list>
# define LL long long
using namespace std ;

char str[50][50];
int p[50][50];
int n,m;
int dfs()
{
    int i,j,flag;
    flag = 0;
    for(i = 0;i < n-1;i ++)
    {
        for(j = 0;j < m-1;j ++)
        {
            if(p[i][j] == 0&&p[i+1][j] == 0&&p[i][j+1] == 0&&p[i+1][j+1] == 0)
            {
                p[i][j] = p[i+1][j] = p[i][j+1] = p[i+1][j+1] = 1;
                if(dfs() == 0) //子状态存在必败态
                flag = 1;
                p[i][j] = p[i+1][j] = p[i][j+1] = p[i+1][j+1] = 0;
            }
        }
    }
    if(flag)
    return 1;
    else
    return 0;
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i = 0; i < n; i ++)
            scanf("%s",str[i]);
        for(i = 0; i < n; i ++)
        {
            for(j = 0; j < m; j ++)
            {
                if(str[i][j] == '0')
                p[i][j] = 0;
                else
                p[i][j] = 1;
            }
        }
        if(dfs())
        printf("Yes
");
        else
        printf("No
");
    }
    return 0;
}