树形DP URAL 1039 Anniversary Party

题目传送门

 /*
     题意：上司在，员工不在，反之不一定。每一个人有一个权值，问权值和最大多少。
     树形DP：把上司和员工的关系看成根节点和子节点的关系，两者有状态转移方程：
             dp[rt][0] += max (dp[son][1], dp[son][0]);    //上司不去
             dp[rt][1] += dp[son][0];                    //上司去，员工都不去
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 using namespace std;

 const int MAXN = 6e3 + ;
 const int INF = 0x3f3f3f3f;
 bool vis[MAXN];
 vector<int> edge[MAXN];
 int dp[MAXN][];
 int n;

 void DFS(int rt)
 {
     vis[rt] = true;
     dp[rt][] = ;
     for (int i=; i<edge[rt].size (); ++i)
     {
         int son = edge[rt][i];
         if (!vis[son])
         {
             DFS (son);
             dp[rt][] += max (dp[son][], dp[son][]);
             dp[rt][] += dp[son][];
         }
     }
 }

 int main(void)        //URAL 1039 Anniversary Party
 {
 //    freopen ("URAL_1039.in", "r", stdin);

     while (scanf ("%d", &n) == )
     {
         memset (vis, false, sizeof (vis));
         memset (dp, , sizeof (dp));
         for (int i=; i<=n; ++i)    scanf ("%d", &dp[i][]);

         int l, k;
         while (scanf ("%d%d", &l, &k) == )
         {
             if (l ==  && k == )    break;
             vis[l] = true;
             edge[l].push_back (k);
             edge[k].push_back (l);
         }

         int root = ;
         for (int i=; i<=n; ++i)
         {
             if (!vis[i])    {root = i;    break;}
         }

         memset (vis, false, sizeof (vis));    DFS (root);
         printf ("%d\n", max (dp[root][], dp[root][]));
     }

     return ;
 }