solution：按行逐个摆，用vis标记第i行有没有棋子就完事了

#include <iostream>
#include <algorithm>
using namespace std;

int n, k, vis[8], cnt;
char maze[8][8];

void dfs(int x, int y)
{
	for (int i = 0; i < n; ++i)
	{
		if (maze[x][i] == '#' && !vis[i]){
			if (y == 1)++cnt;
			else{
				vis[i] = 1;
				for (int j = x + 1; j < n - y + 2; ++j)dfs(j, y - 1);
				vis[i] = 0;
			}
		}
	}
}

int main()
{
	while (cin >> n >> k)
	{
		if (n == -1 && k == -1)break;
		fill(vis, vis + n, 0);
		for (int i = 0; i < n; ++i){
			cin >> maze[i];
		}
		cnt = 0;
		for (int i = 0; i <= n; ++i){
			dfs(i, k);
		}
		cout << cnt << endl;
	}
	return 0;
}