PKU_campus_2017_K Lying Island

思路：

题目链接http://poj.openjudge.cn/practice/C17K/

状压dp。dp[i][j]表示第i - k人到第i人的状态为j的情况下前i人中最多有多少好人。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 int dp[][];
 struct node
 {
     int type, id1, id2;
     bool f1, f2;
 };
 node a[];
 bool same(bool a, bool b)
 {
     return (a && b) || (!a && !b);
 }
 int main()
 {
     int t, n, k;
     cin >> t;
     while (t--)
     {
         memset(dp, , sizeof dp);
         cin >> n >> k;
         string s;
         getchar();
         for (int i = ; i < n; i++)
         {
             getline(cin, s);
             string tmp;
             stringstream ss(s);
             vector<string> v;
             while (ss >> tmp) v.push_back(tmp);
             if (v[][] == 'I')
             {
                 a[i].type = ;
                 a[i].id1 = atoi(v[].c_str());
                 a[i].id2 = atoi(v[].c_str());
                 a[i].f1 = v[] == "good" ? true : false;
                 a[i].f2 = v[] == "good" ? true : false;
             }
             else
             {
                 a[i].type = ;
                 a[i].id1 = atoi(v[].c_str());
                 a[i].f1 = v[] == "good" ? true : false;
             }
         }

         int msk = ( << k + ) - ;
         dp[][] = ;
         for (int i = ; i < n; i++)
         {
             memset(dp[i & ], , sizeof dp[i & ]);
             for (int j = ; j <  << k + ; j++)
             {
                 if (i < k +  && j >=  << i) continue;
                 int tmp = j <<  & msk;
                 dp[i & ][tmp] = max(dp[i & ][tmp], dp[i -  & ][j]);
                 if (a[i].type == )
                 {
                     int p1 = i - a[i].id1;
                     if (same(a[i].f1, j >> p1 & ))
                     {
                         dp[i & ][tmp | ] = max(dp[i & ][tmp | ], dp[i -  & ][j] + );
                     }
                 }
                 else
                 {
                     int p1 = i - a[i].id1, p2 = i - a[i].id2;
                     bool g1 = j >> p1 & , g2 = j >> p2 & ;
                     if (!(same(a[i].f1, g1) && !same(a[i].f2, g2)))
                         dp[i & ][tmp | ] = max(dp[i & ][tmp | ], dp[i -  & ][j] + );
                 }
             }
         }
         int ans = ;
         for (int i = ; i <  << k + ; i++)
             ans = max(ans, dp[n -  & ][i]);
         cout << ans << endl;
     }
     return ;
 }