https://www.acwing.com/problem/content/1075/

輸出樹的中心（該點到樹中其他結點的最遠距離最近）。

時間復雜度 \(O(n)\)。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	int n;
	cin >> n;
	vector < vector < array<LL, 2> > > e(n + 1);
	for (int i = 1; i < n; i ++ ){
		int u, v, w;
		cin >> u >> v >> w;
		e[u].push_back({w, v});
		e[v].push_back({w, u});
	}
	vector <LL> d1(n + 1), d2(n + 1), s1(n + 1), s2(n + 1);
	function<void(int, int)> dfs1 = [&](int u, int fa){
		for (auto [w, v] : e[u]){
			if (v == fa) continue;
			dfs1(v, u);
			LL x = d1[v] + w;
			if (x > d1[u]){
				d2[u] = d1[u], s2[u] = s1[u];
				d1[u] = x, s1[u] = v;
			}
			else if (x > d2[u]){
				d2[u] = x, s2[u] = v;
			}
		}
	};
	dfs1(1, 0);
	vector <LL> up(n + 1);
	function<void(int, int)> dfs2 = [&](int u, int fa){
		for (auto [w, v] : e[u]){
			if (v == fa) continue;
			if (s1[u] == v) up[v] = max(up[u], d2[u]) + w;
			else up[v] = max(up[u], d1[u]) + w;
			dfs2(v, u);
		}
	};
	dfs2(1, 0);
	LL ans = 1e18;
	for (int i = 1; i <= n; i ++ )
		ans = min(ans, max(d1[i], up[i]));
	cout << ans << "\n";
	return 0;
}