hihocoder编程练习赛52-2 亮灯方案

思路：

状态压缩dp。
实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int MAXN = ;
 const int dx[] = {, , -, };
 const int dy[] = {, , , -};
 int n, m;
 ll dp[( << ) + ];
 char a[MAXN][MAXN], tmp[MAXN][MAXN];
 bool check(int x, int y, int S)
 {
     for (int i = ; i < ; i++)
     {
         int nx = x + dx[i], ny = y + dy[i];
         if (nx >=  && nx < n && ny >=  && ny < m && (S & ( << (nx * m + ny))))
             return true;
     }
     return false;
 }
 ll dfs(int now, int S)
 {
     if (dp[S] != -) return dp[S];
     if (now == n * m) return ;
     ll cnt = ;
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             int x = i * m + j;
             if (!(S & ( << x)) && check(i, j, S))
                 cnt += dfs(now + , S | ( << (i * m + j)));
         }
     }
     return dp[S] = cnt;
 }

 int main()
 {
     cin >> n >> m;
     int now = , S = ;
     for (int i = ; i < n; i++)
         for (int j = ; j < m; j++)
             cin >> a[i][j];
     for (int i = ; i < n; i++)
     {
         for (int j = ; j < m; j++)
         {
             if (a[i][j] == '')
             {
                 now++;
                 S |= ( << i * m + j);
             }
         }
     }
     memset(dp, -, sizeof dp);
     cout << dfs(now, S) << endl;
     return ;
 }