算法1：两次BFS

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int n, u, v, w;

pair<int, long> bfs(vector<vector<pair<int, int>>>& g, int s) {
  queue<pair<int, int>> q;
  q.emplace(s, 0);
  vector<int> seen(n + 1);
  seen[s] = 1; // mark as seen

  pair<int, long> ans{-1, 0};
  while (not q.empty()) {
    const auto [u, d] = q.front(); q.pop();
    if (d > ans.second) ans.first = u, ans.second = d;
    for (const auto& [v, w] : g[u]) {
      if (seen[v]++) continue;
      q.emplace(v, d + w);
    }
  }

  return ans;
};

// debug helper function
void printGraph(vector<vector<pair<int, int>>>& g) {
  for (int u = 1; u < g.size(); ++u) {
    printf("%d's neighbours: [", u);
    for (const auto& [v, w] : g[u])
      printf("(%d, %d) ", v, w);
    printf("]\n");
  }
}

int main(void) {
  scanf("%d", &n);
  // build the undirected graph
  vector<vector<pair<int, int>>> g(n + 1); // 1-Based
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d %d", &u, &v, &w); // w == weight (权重)
    g[u].emplace_back(v, w);
    g[v].emplace_back(u, w);
  }
  // printGraph(g);
  long dist = bfs(g, bfs(g, 1).first).second;
  printf("%ld\n", dist * 10 + ((dist + 1) * dist >> 1));
}

TODO： 算法2：两次DFS