树的最长直径（邻接表）

可以证明从树的任意一点出发，所能到达的最远距离一定是树的最大直径的端点。所以两次bfs(),一次任意一点出发，一次端点出发，便能找到最大直径。

程序代码如下：（求树上两端点的最远距离）

#include <iostream>
#include <string.h>
#define MAXN 10005
#define INF 999999999

using namespace std;

struct Edge
{
    int v, w, next;
}edge[MAXN];

int head[MAXN], vis[MAXN], d[MAXN], q[MAXN], e, n;

int init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w)
{
    edge[e].v = v;
    edge[e].w = w;
    edge[e].next = head[u];
    head[u] = e++;
}

void bfs(int src)
{
    for （int i = 0; i <= n; i++)
    {
        vis[i] = 0;
        d[i] = INF;
    }

    int h = 0, t = 0;
    vis[src] = 1;
    q[t++] = src;
    d[src] = 0;
    while (h < t)
    {
        int u = q[h++];
        for (int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            int w = edge[i].w;
            if (d[u] + w < d[v])
            {
                d[v] = d[u] + w;
                if (!vis[v])
                {
                    q[t++] = v;
                    vis[v] = 1;
                }
            }
        }
    }
}

int main()
{
    int u, v, w;
    init();
    cin >> n;
    for (int i = 1; i <= n - 1; i++)
    {
        cin >> u >> v >> w;
        add(u, v, w);
        add(v, u, w);
    }
    bfs(1);
    int pos = 0, mx = -1;
    for (int i  = 1; i <= n; i++)
    {
        if(d[i] > mx)
        {
            mx = d[i];
            pos = i;
        }
    }
    bfs(pos);
    mx = -1;
    for (int i  = 1; i <= n; i++)
    {
        if(d[i] > mx)
        {
            mx = d[i];
        }
    }
    cout << mx << endl;
    return 0;
}