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