D - Fliptile

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <string.h>
 7 using namespace std;
 8 
 9 const int INF = 0xfffffff;
10 int g[17][17];
11 int f[17][17] = {};
12 int ans[17][17] ={};
 13  int mmin = INF;
 14  
15  bool judge( int n, int m) // judge whether the last line is all 0 
16  {
 17      for ( int i= 1 ;i<=m;i++ )
 18      {
 19          int t = f[n][i]+f[n][i- 1 ]+f[n][i+ 1 ]+f[n- 1 ][i];
 20          if ( (g[n][i] +t)& 1 )
 21              return  false ;
 22          }
 23      return  true;
 24  }
 25  
26  void dfs( int n, int m, int k, int num)
 27  {
 28      if (num > mmin) // prune 
29          return ;
 30      if (k > n)
 31      {
 32          if (judge( n, m) && mmin>num) // Determine whether the condition is met 
33          {
 34              memcpy(ans, f, sizeof (f));
 35              mmin = num;
 36          }
 37         return ;
 38      }
 39      int t = 0 ;
 40      for ( int i= 1 ;i<=m;i++ )
 41      {
 42          if ((g[k- 1 ][i]+f[k- 2 ][i] +f[k- 1 ][i- 1 ]+f[k- 1 ][i+ 1 ]+f[k- 1 ][i])& 1 ) // Whether the previous line is 1, that is, whether it needs to flip 
43          {
 44              f[k][i] = 1 ;
 45              t++ ;
 46         }
 47          else 
48              f[k][i] = 0 ;
 49      }
 50      dfs(n, m, k+ 1 , num+ t);
 51  }
 52  
53  // n,m row and column number k The first row of the current column num is flipped Times 
54  void todfs ( int n, int m, int k, int num)
 55  {
 56      if (k > m)
 57      {
 58          dfs(n, m, 2 , num); // For each case of the first row Search 
59          return;
 60      }
 61      f[ 1 ][k] = 0 ;       // do not flip 
62      todfs(n, m, k+ 1 , num);
 63      f[ 1 ][k] = 1 ;       // flip, num+1 
64      todfs( n, m, k+ 1 , num+ 1 );
 65  }
 66  
67  int main()
 68  {
 69      int n,m;
 70      cin >> n >> m;
 71      for ( int i= 1;i<=n;i++ )
 72          for ( int j= 1 ;j<=m;j++ )
 73              cin>> g[i][j];
 74      todfs(n, m, 1 , 0 );      // recursion Iterate over all cases in the first line 
75      if (mmin == INF)
 76          cout<< " IMPOSSIBLE " << endl;
 77      else 
78          for ( int i= 1 ;i<=n;i++ )
 79          {
 80              for ( int j=1;j<=m;j++)
81                 cout<<ans[i][j]<<" ";
82             cout<<endl;
83         }
84     return 0;
85 }