树形DP Gym 100496H House of Representatives

题目传送门

 /*
     题意：寻找一个根节点，求min f(u) = ∑ρ(v, u) * p(v)。ρ(v, u)是u到v的距离，p(v)是v点的权值
     树形DP：先从1出发遍历第一次，sum[u]计算u到所有子节点v的路径权值(之后的点路径有叠加，所以先把路径权值加后*w)，
             计算f[u](缺少u节点以上的信息)。然后再遍历一遍，之前是DFS从下往上逆推，现在是顺推，把u节点以上的信息加上
     dp的部分不是很多，两个DFS函数想了很久，还是没完全理解:(
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int p[MAXN];
 ll f[MAXN];
 ll sum[MAXN];
 vector<pair<int, int> > G[MAXN];

 void DFS1(int u, int rt)
 {
     f[u] = ;    sum[u] = p[u];
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i].first;    int w = G[u][i].second;
         if (v == rt)    continue;
         DFS1 (v, u);
         f[u] += f[v] + sum[v] * w;    sum[u] += sum[v];
     }
 }

 void DFS2(int u, int rt)
 {
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i].first;    int w = G[u][i].second;
         if (v == rtGym 100496H House of Representatives)    continue;
         f[v] = f[u] - sum[v] * w + (sum[u] - sum[v]) * w;
         sum[v] = sum[u];
         DFS2 (v, u);
     }
 }

 int main(void)        //Gym 100496H House of Representatives
 {
 //    freopen ("H.in", "r", stdin);
     freopen ("house.in", "r", stdin);
     freopen ("house.out", "w", stdout);

     int n;
     while (scanf ("%d", &n) == )
     {
         for (int i=; i<=n; ++i)    scanf ("%d", &p[i]);
         for (int i=; i<=n; ++i)    G[i].clear ();

         for (int i=; i<=n-; ++i)
         {
             int u, v, w;    scanf ("%d%d%d", &u, &v, &w);
             G[u].push_back (make_pair (v, w));
             G[v].push_back (make_pair (u, w));
         }

         DFS1 (, -);    DFS2 (, -);

         int p = ;
         for (int i=; i<=n; ++i)
         {
             if (f[i] < f[p])    p = i;
         }

         printf ("%d %I64d\n", p, f[p]);
     }

     return ;
 }