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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-1lines will contain three integers u v w (0 ≤ u, v < n, u ≠ v, 1 ≤ w ≤ 10000)

Output

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

Notes

Dataset is huge, use faster i/o methods.

先一次搜索搜索到最遠的端點。然後再從這個端點搜索整個樹。即為樹中最遠的距離，就是樹的直徑

PROBLEM SETTER: JANE ALAM JAN

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
#define M 31000
struct node {
	int v,next,w;
}mp[M*3];
int cnt,head[M],dis[M],vis[M];
int ans,last;
void add(int u,int v,int w){
	mp[cnt].v=v;
	mp[cnt].w=w;
	mp[cnt].next=head[u];
	head[u]=cnt++;
}
void bfs(int s){
	queue<int> q;
	memset(vis,0,sizeof(vis));
	memset(dis,0,sizeof(dis));
	vis[s]=1;
	last=s;
	ans=0;
	q.push(s);
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=mp[i].next){
			int v=mp[i].v;
			if(!vis[v] && dis[v]<dis[u]+mp[i].w){
				vis[v]=1;
				dis[v]=dis[u]+mp[i].w;
				if(ans<dis[v]){
					ans=dis[v];
					last=v;
				}
				q.push(v);
			}
		}
	}
}
int main(){
	int t,n,a,b,c,k=1;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		cnt=0;
		memset(head,-1,sizeof(head));
		for(int i=0;i<n-1;i++){
			scanf("%d %d %d",&a,&b,&c);
			add(a,b,c);
			add(b,a,c);
		}
		bfs(0);
		bfs(last);
		printf("Case %d: ", k++); 
		printf("%d\n",ans); 
	}
	return 0;
} 

light oj 1094 Farthest Nodes in a Tree（樹的直徑模板）