牛客-Highway

题目传送门

sol:看了题意显然是最大生成树，但是任意两个点之间都有边，大概有n*n条边。用朴素的最小生成树算法显然不行。联想了一下树的直径还是不会。看了大佬的题解，懂了。。。

所以还是直接贴大佬博客链接好了：https://blog.csdn.net/yasola/article/details/72229734

• 树的直径

  #include "bits/stdc++.h"
  using namespace std;
  typedef long long LL;
  typedef pair<int, int> PII;
  typedef pair<int, LL> PIL;
  const int MAXN = 1e5 + ;
  vector<PII> edge[MAXN];
  LL _max; int nn1, nn2;
  LL dis[MAXN];
  PIL get_diameter(int rt, int fa) {
      LL m1 = , m2 = ;
      int n1 = rt, n2 = rt;
      for (PII i : edge[rt]) {
          if (i.first == fa) continue;
          PIL p = get_diameter(i.first, rt);
          if (p.second + i.second > m1) {
              m2 = m1, n2 = n1;
              m1 = p.second + i.second;
              n1 = p.first;
          } else if (p.second + i.second > m2) {
              m2 = p.second + i.second;
              n2 = p.first;
          }
      }
      if (m1 + m2 > _max) {
          nn1 = n1, nn2 = n2;
          _max = m1 + m2;
      }
      return {n1, m1};
  }
  void dfs(int rt, int fa, LL mm) {
      dis[rt] = max(dis[rt], mm);
      for (PII i : edge[rt]) {
          if (i.first == fa) continue;
          dfs(i.first, rt, mm + i.second);
      }
  }
  int main() {
      int n;
      while (~scanf("%d", &n)) {
          memset(dis, -, sizeof(dis));
          for (int i = ; i <= n; i++) edge[i].clear();
          for (int i = ; i < n; i++) {
              int u, v, w;
              scanf("%d%d%d", &u, &v, &w);
              edge[u].push_back({v, w});
              edge[v].push_back({u, w});
          }
          _max = -;
          get_diameter(, -);    // 求出树的直径，以及两个端点；
          dfs(nn1, -, ), dfs(nn2, -, );
          LL sum = ;
          for (int i = ; i <= n; i++) sum += dis[i];
          printf("%lld\n", sum - _max);    // 将两个端点加入集合只用算一次直径，而上面的循环算了两次，所以减掉一个直径；
      }
      return ;
  }