Codeforces 294E Shaass the Great

树形DP。由于n只有5000，可以直接枚举边。

枚举边，将树分成两个子树，然后从每个子树中选出一个点分别为u,v，那么答案就是:

子树1中任意两点距离总和+子树2中任意两点距离总和+子树1中任意一点到u的距离和*子树2的节点个数+子树2中任意一点到v的距离和*子树1的节点个数+子树1的节点个数*子树2的节点个数*当前枚举边的权值。

当枚举的边一定时，那么要选取的点就是子树中到所有点的距离总和最小的点。对树进行dfs，同时记录子树的节点个数，所有孩子节点到当前根节点的距离总和，以及当前子树中任意两点距离和。然后在进行dfs求解到该子树中所有点距离和最小的点。

 #include <stdio.h>
 #include <string.h>
 #include <vector>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> pii;
 #define maxn 5005
 #define INF 0x3f3f3f3f3f3f3f3fll

 int f[maxn], h[maxn], w[maxn];
 vector<pii> g[maxn];
 int son[maxn];
 LL d[maxn], s[maxn];

 void dfs(int v, int rt){
     son[v] = , d[v] = , s[v] = ;
     for(int i = ; i != g[v].size(); i ++){
         int u = g[v][i].first;
         if(u==rt) continue;
         dfs(u, v);
         s[v] += s[u] + d[u]*son[v] + d[v]*son[u] + (LL)son[u] * son[v] * g[v][i].second;
         d[v] += d[u] + (LL)son[u] * g[v][i].second;
         son[v] += son[u];
     }
     son[v] ++;
     s[v] += d[v];
 }
 void dfsw(int v, int rt, LL &minn){
     minn = min(d[v], minn);
     for(int i = ; i != g[v].size(); i ++){
         int u = g[v][i].first;
         if(u==rt) continue;
         d[u] = d[u] + (d[v] - d[u] - (LL)son[u]*g[v][i].second) + (LL)(son[v] - son[u])*g[v][i].second;
         son[u] = son[v];
         dfsw(u, v, minn);
     }
 }
 int main(){
     //freopen("test.in", "r", stdin);
     for(int n; scanf("%d", &n)!=EOF; ){
         for(int i = ; i <= n; i ++){
             g[i].clear();
         }
         for(int i = , x, y, z; i < n; i ++){
             scanf("%d%d%d", &x, &y, &z);
             f[i] = x, h[i] = y, w[i] = z;
             g[x].push_back(make_pair(y, z));
             g[y].push_back(make_pair(x, z));
         }
         LL ans = INF, minn = INF, sum = ;
         for(int i = ; i < n; i ++){
             minn = INF;
             dfs(f[i], h[i]);
             dfsw(f[i], h[i], minn);
             sum = ;
             sum = (n - son[f[i]]) * minn + s[f[i]];
             minn = INF;
             dfs(h[i], f[i]);
             dfsw(h[i], f[i], minn);
             sum = sum + (n - son[h[i]]) * minn + s[h[i]] + (LL)son[f[i]]*son[h[i]]*w[i];
             ans = min(ans, sum);
         }
         printf("%I64d\n", ans);
     }
     return ;
 }