poj 2631 Roads in the North【树的直径裸题】

Roads in the North

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2359		Accepted: 1157

Description

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice. 
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area. 

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1. 

Input

Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way.

Output

You are to output a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 6
1 4 5
6 3 9
2 6 8
6 1 7

Sample Output

22

告诉我不是我一个人看到输入数据时不知道是什么意思
树的直径裸题

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 40000
using namespace std;
int head[MAX];
int vis[MAX],dis[MAX];
int ans,sum,beg;
struct node
{
	int u,v,w;
	int next;
}edge[MAX];
void add(int u,int v,int w)
{
	edge[ans].u=u;
	edge[ans].v=v;
	edge[ans].w=w;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void bfs(int sx)
{
	int i,j;
	memset(vis,0,sizeof(vis));
	memset(dis,0,sizeof(dis));
	queue<int>q;
	sum=0;
	beg=sx;
	q.push(sx);
	while(!q.empty())
	{
	    int top=q.front();
            q.pop();
		for(i=head[top];i!=-1;i=edge[i].next)
		{
			int x=edge[i].v;
			if(!vis[x])
			{
				
				vis[x]=1;
				dis[x]=dis[top]+edge[i].w;
				q.push(x);
				if(sum<dis[x])
				{
					sum=dis[x];
					beg=x;
				}
			}
		}	 
	}
}
int main()
{
	ans=0;
	memset(head,-1,sizeof(head));
	int a,b,c;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		add(a,b,c);
		add(b,a,c);
	}
	bfs(1);
	bfs(beg);
	printf("%d\n",sum);
	return 0;
}