Hdu 1760 dfs+ Game

0 for the 1 with the table can not put a 2*2 each time the block can not let the loser
Output Yes if the initiator is winning

Winning state: There is a must-have in the status that can be reached from the current state
The state that can be achieved from the current state is all a winning

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

Sample Output
Yes
No

1# include <iostream>2# include <cstdio>3# include <cstring>4# include <algorithm>5# include <string>6# include <cmath>7# include <queue>8# include <list>9# define LLLong LongTen using namespacestd; One  A Charstr[ -][ -]; - intp[ -][ -]; - intn,m; the intDFS () - { -     intI,j,flag; -Flag =0; +      for(i =0; i < n-1; i + +) -     { +          for(j =0; J < M1; j + +) A         { at             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)//There must be a state of sub-states -Flag =1; -P[I][J] = p[i+1][J] = p[i][j+1] = p[i+1][j+1] =0; in             } -         } to     } +     if(flag) -     return 1; the     Else *     return 0; $ }Panax Notoginseng intMain () - { the     inti,j; +      while(SCANF ("%d%d", &n,&m)! =EOF) A     { the          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') theP[I][J] =0; -                 ElseWuyiP[I][J] =1; the             } -         } Wu         if(Dfs ()) -printf"yes\n"); About         Else $printf"no\n"); -     } -     return 0; -}

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Hdu 1760 dfs+ Game