法一

模板

// O(n)

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 10010, M = 20010;

int n, u;

int q[N];
int dis[N];
int h[N], e[M], ne[M], w[M], idx;

void add (int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++ ;
}

int bfs (int x)
{
    memset(dis, -1, sizeof dis);
    dis[x] = 0;

    int hh = 0, tt = 0;
    q[0] = x;

    while(hh <= tt)
    {
        int t = q[hh ++ ];

        for(int i = h[t] ; ~i ; i = ne[i])
        {
            int j = e[i];
            if(dis[j] == -1)
            {
                dis[j] = dis[t] + w[i];
                q[++ tt] = j;
            }
        }
    }

    if(x == 1)
    {
        int t = 1;
        for(int i = 2 ; i <= n ; i ++ )
            if(dis[i] > dis[t])
                t = i;

        return t;
    }

    if(x == u)
    {
        int mmax = 0;
        for(int i = 1 ; i <= n ; i ++ ) mmax = max(mmax, dis[i]);
        return mmax;
    }
}

int main ()
{
    cin >> n;

    memset(h, -1, sizeof h);
    for(int i = 0 ; i < n - 1 ; i ++ )
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }

    u = bfs(1);
    if(u == 1) // 这样传上去的话答案会走第一个if语句
    {
        puts("0");
        return 0;
    }
    cout << bfs(u) << endl;

    return 0;
}

法二

树形dp
还未学待更新....