题目大意：给出一棵树。求两点间的最长距离。

思路：裸地树的直径。两次BFS，第一次随便找一个点宽搜。然后用上次宽搜时最远的点在宽搜。得到的最长距离就是树的直径。

CODE：

#include <queue>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#define MAX 80010

using namespace std;

int points,edges;

int head[MAX],total;

int next[MAX],aim[MAX],length[MAX];

int f[MAX];

char s[10];

inline void Add(int x,int y,int len);

inline void BFS(int start);

int main()

{

	cin >> points >> edges;

	for(int x,y,len,i = 1;i <= edges; ++i) {

		scanf("%d%d%d%s",&x,&y,&len,s);

		Add(x,y,len),Add(y,x,len);

	}

	BFS(1);

	int _max = 0;

	for(int i = 1;i <= points; ++i)

		if(f[i] > f[_max])

			_max = i;

	BFS(_max);

	int ans = 0;

	for(int i = 1;i <= points; ++i)

		if(f[i] > ans)

			ans = f[i];

	cout << ans << endl;

	return 0;

}

inline void Add(int x,int y,int len)

{

	next[++total] = head[x];

	aim[total] = y;

	length[total] = len;

	head[x] = total;

}

inline void BFS(int start)

{

	static queue<int> q;

	while(!q.empty())	q.pop();

	memset(f,0x3f,sizeof(f));

	f[0] = f[start] = 0;

	q.push(start);

	while(!q.empty()) {

		int x = q.front(); q.pop();

		for(int i = head[x];i;i = next[i])

			if(f[aim[i]] > f[x] + length[i]) {

				f[aim[i]] = f[x] + length[i];

				q.push(aim[i]);

			}

	}

}