poj 2631 Roads in the North (树的直径)

poj2631 题意：一棵树中求两个点的最远距离。。。 思路：就是求树的直径。。。裸体。。。。1A

/* ***********************************************
Author :111qqz
Created Time :2016年07月12日 星期二 13时03分39秒
File Name :code/poj/2631.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E4+7;
int n,m;
vector < pi> edge[N];
int lst;
int ans;
int d[N];
bool vis[N];


void bfs( int s)
{
    ms(d,0x3f);
    ms(vis,false);
    queue<int>q;
    q.push(s);
    vis[s] = true;
    d[s] =  0 ;

    while (!q.empty())
    {
	int u = q.front() ; q.pop();
	lst = u ;
//	cout<<"u:"<<u<<endl;
	int siz = edge[u].size();

	for ( int i = 0 ; i < siz ; i++)
	{
	    int v = edge[u][i].fst;
	    if (vis[v]) continue;
	    vis[v] = true;
	    d[v] = d[u] + edge[u][i].sec;
	    ans = max(d[v],ans);
	    q.push(v);
	}
    }

    int mx = 0;
    for ( int i = 1 ; i <= n ; i++)
    {
//	cout<<"i:"<<i<<" d[i]:"<<d[i]<<endl;
	if (d[i]>mx)
	{
	    mx = d[i];
	    lst = i ;
	}
    }

}
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	int u,v,w;
	m = 0;
	n = 0;
	while (scanf("%d%d%d",&u,&v,&w)!=EOF)
	{
	    edge[u].push_back(make_pair(v,w));
	    edge[v].push_back(make_pair(u,w));
	    m++;
	    n = max(n,u);
	    n = max(n,v);
	}
	ans = inf;
	bfs(1);
	ans = 0;
//	cout<<"lst:"<<lst<<endl;
	bfs(lst);
	cout<<ans<<endl;

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}