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[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
char a[1005][1005], b[1005][1005];
int ok[1005][1005];
int main()
{
    int n, m;
    while (cin >> n >> m)
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                cin >> a[i][j];
                if (i == 0 || i == n - 1 || j == 0 || j == m - 1)
                    ok[i][j] = 0;
                else ok[i][j] = 1;
            }
        }
        for (int i = 1; i < n - 1; i++)
        {
            for (int j = 1; j < m - 1; j++)
            {
                if (a[i][j] == '.')
                {
                    for (int k = 0; k < 8; k++)
                    {
                        int nx = i + dx[k], ny = j + dy[k];
                        ok[nx][ny] = 0;
                    }
                }
            }
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (ok[i][j])
                {
                    for (int k = 0; k < 8; k++)
                    {
                        int nx = i + dx[k], ny = j + dy[k];
                        b[nx][ny] = '#';
                    }
                }
            }
        }
        bool flg = true;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (a[i][j] == '#' && b[i][j] != '#')
                {
                    flg = false; break;
                }
            }
        }
        cout << (flg ? "yes" : "no") << endl;
    }
    return 0;
}