CF1059B Forgery

思路：

若某个位置是‘.’，说明不能在周围的8个位置下笔。在所有可以下笔的位置填充一次，看能否“包含”需要的图案即可。

实现：

 #include <iostream>
 using namespace std;
 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {, , , -, , -, , -};
 char a[][], b[][];
 int ok[][];
 int main()
 {
     int n, m;
     while (cin >> n >> m)
     {
         for (int i = ; i < n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 cin >> a[i][j];
                 if (i ==  || i == n -  || j ==  || j == m - )
                     ok[i][j] = ;
                 else ok[i][j] = ;
             }
         }
         for (int i = ; i < n - ; i++)
         {
             for (int j = ; j < m - ; j++)
             {
                 if (a[i][j] == '.')
                 {
                     for (int k = ; k < ; k++)
                     {
                         int nx = i + dx[k], ny = j + dy[k];
                         ok[nx][ny] = ;
                     }
                 }
             }
         }
         for (int i = ; i < n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 if (ok[i][j])
                 {
                     for (int k = ; k < ; k++)
                     {
                         int nx = i + dx[k], ny = j + dy[k];
                         b[nx][ny] = '#';
                     }
                 }
             }
         }
         bool flg = true;
         for (int i = ; i < n; i++)
         {
             for (int j = ; j < m; j++)
             {
                 if (a[i][j] == '#' && b[i][j] != '#')
                 {
                     flg = false; break;
                 }
             }
         }
         cout << (flg ? "yes" : "no") << endl;
     }
     return ;
 }