LightOJ--1094-- Farthest Nodes in a Tree(树的直径裸题）

Farthest Nodes in a Tree

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

Description

Given a tree (a connected graph with no cycles), you have to find the farthest nodes in the tree. The edges of the tree are weighted and undirected. That means you have to find two nodes in the tree whose distance is maximum amongst all nodes.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with an integer n (2 ≤ n ≤ 30000) denoting the total number of nodes in the tree. The nodes are numbered from 0 to n-1. Each of the next n-1 lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000) denoting that node u and v are connected by an edge whose weight is w. You can assume that the input will form a valid tree.

Output

For each case, print the case number and the maximum distance.

Sample Input

2

4

0 1 20

1 2 30

2 3 50

5

0 2 20

2 1 10

0 3 29

0 4 50

Sample Output

Case 1: 100

Case 2: 80

Hint

Source

Problem Setter: Jane Alam Jan 

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAXN 30000+10
#define MAXM 900000+10
#define INF 0x3f3f3f
int n,cnt,head[MAXN],vis[MAXN],dis[MAXN];
int ans,sx;
struct node
{
	int u,v;
	int val,next;
}edge[MAXM];
void add(int u,int v,int val)
{
	node E={u,v,val,head[u]};
	edge[cnt]=E;
	head[u]=cnt++;
}
void bfs(int x)
{
	queue<int>q;
	ans=0;
	memset(vis,0,sizeof(vis));
	memset(dis,0,sizeof(dis));
	q.push(x);
	vis[x]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			node E=edge[i];
			if(!vis[E.v]&&dis[E.v]<dis[u]+E.val)
			{
				vis[E.v]=1;
				dis[E.v]=dis[u]+E.val;
				q.push(E.v);
				if(dis[E.v]>ans)
				{
					ans=dis[E.v];
					sx=E.v;
				}
			}
		}
	}
}
int main()
{
	int t;
	int k=1;
	cin>>t;
	while(t--)
	{
		cin>>n;
		cnt=0;
		memset(head,-1,sizeof(head));
		int a,b,c;
		for(int i=1;i<n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			a++,b++;
			add(a,b,c);
			add(b,a,c);
		}
		sx=1;
		bfs(1);
		bfs(sx);
		printf("Case %d: %d\n",k++,ans);
	}
	return 0;
}