poj3279 Fliptile

思路：

枚举。

枚举了第一行的操作之后，下面每行的操作也随之确定了。因为在确定了第i行的操作之后，要想再改变a[i][j]的状态只能通过改变a[i + 1][j]来实现。另外，用到了集合的整数表示方法。

实现：

 #include <iostream>
 #include <cstring>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 const int dx[] = {, , , -};
 const int dy[] = {, , -, };
 int a[][], b[][], ans[][], buf[][], m, n;
 void turn(int x, int y)
 {
     b[x][y] =  - b[x][y];
     for (int i = ; i < ; i++)
     {
         int nx = x + dx[i], ny = y + dy[i];
         if (nx >=  && nx < m && ny >=  && ny < n)
         {
             b[nx][ny] =  - b[nx][ny];
         }
     }
 }
 int main()
 {
     cin >> m >> n;
     for (int i = ; i < m; i++)
     {
         for (int j = ; j < n; j++)
             cin >> a[i][j];
     }
     int minn = INF;
     for (int i = ; i < ( << n); i++)
     {
         int cnt = ;
         memset(buf, , sizeof buf);
         for (int i = ; i < m; i++)
         {
             for (int j = ; j < n; j++)
             {
                 b[i][j] = a[i][j];
             }
         }
         for (int j = ; j < n; j++)
         {
             int msk =  << j;
             if (i & msk)
             {
                 turn(, j); cnt++; buf[][j] = ;
             }
         }
         for (int j = ; j < m; j++)
         {
             for (int k = ; k < n; k++)
             {
                 if (b[j - ][k])
                 {
                     turn(j, k); cnt++; buf[j][k] = ;
                 }
             }
         }
         bool flg = true;
         for (int i = ; i < n; i++)
         {
             if (b[m - ][i]) { flg = false; break; }
         }
         if (flg && cnt < minn)
         {
             minn = cnt;
             for (int i = ; i < m; i++)
             {
                 for (int j = ; j < n; j++)
                 {
                     ans[i][j] = buf[i][j];
                 }
             }
         }
     }
     if (minn == INF) cout << "IMPOSSIBLE" << endl;
     else
     {
         for (int i = ; i < m; i++)
         {
             for (int j = ; j < n; j++)
                 cout << ans[i][j] << " ";
             cout << endl;
         }
     }
     return ;
 }