hihocoder offer收割编程练习赛11 B 物品价值

思路：

状态压缩 + dp。

实现：

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

 const int INF = 0x3f3f3f3f;

 int t, n, m, s;
 int V[], S[], dp[][( << ) + ];

 int dfs(int now, int s)
 {
     if (dp[now][s] != -)
         return dp[now][s];
     if (now == n)
     {
         if (s == ( << m) - )
             return ;
         return -INF;
     }
     int x = dfs(now + , s ^ S[now]) + V[now];
     int y = dfs(now + , s);
     return dp[now][s] = max(x, y);
 }

 int main()
 {
     cin >> t;
     while (t--)
     {
         memset(dp, -, sizeof(dp));
         cin >> n >> m;
         for (int i = ; i < n; i++)
         {
             cin >> V[i] >> s;
             int tmp = , total = ;
             for (int j = ; j < s; j++)
             {
                 cin >> tmp;
                 tmp--;
                 total |= ( << tmp);
             }
             S[i] = total;
         }
         cout << dfs(, ) << endl;
     }
     return ;
 }