【Light-oj】-1094 - Farthest Nodes in a Tree（树的直径）

最新推荐文章于 2018-09-09 11:26:55 发布

Bear1997

最新推荐文章于 2018-09-09 11:26:55 发布

阅读量416

点赞数

CC 4.0 BY-SA版权

分类专栏：树的直径

本文链接：https://blog.youkuaiyun.com/Bear1998/article/details/52095699

树的直径专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种求解树中两个节点间最大距离的问题，即树的直径算法。通过两次宽度优先搜索找到最远的两个节点及其距离。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1094 - Farthest Nodes in a Tree

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

0 1 20

1 2 30

2 3 50

0 2 20

2 1 10

0 3 29

0 4 50

Case 1: 100

Case 2: 80

Notes

Dataset is huge, use faster i/o methods.

树的直径的裸题，详解看：点击打开题目

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define MAX 40000
int n;
vector<int> link[MAX+10];
vector<int> va[MAX+10]; 
bool vis[MAX+10];
int dis[MAX+10];
void init()
{
	for(int i=0;i<n;i++)
	{
		link[i].clear();
		va[i].clear();
	}
}
int bfs(int x)
{
	int ans;
	int maxx=0;
	memset(vis,false,sizeof(vis));
	memset(dis,0,sizeof(dis));
	queue<int> q;
	dis[x]=0;
	q.push(x);
	vis[x]=true;
	while(!q.empty())
	{
		int st=q.front();
		q.pop();
		for(int i=0;i<link[st].size();i++)
		{
			if(!vis[link[st][i]])
			{
				q.push(link[st][i]);
				dis[link[st][i]]=dis[st]+va[st][i];
				if(maxx<dis[link[st][i]])
				{
					maxx=dis[link[st][i]];
					ans=link[st][i];
				}
				vis[link[st][i]]=true;
			}
		}
	}
	return ans;
}
int main()
{
	int u,cas=1;
	scanf("%d",&u);
	while(u--)
	{
		init();
		scanf("%d",&n);
		for(int i=1;i<n;i++)
		{	
			int t1,t2,t3;
			scanf("%d %d %d",&t1,&t2,&t3);
			link[t1].push_back(t2);
			va[t1].push_back(t3);
			link[t2].push_back(t1);
			va[t2].push_back(t3);
		}
		int z=bfs(0);
		int ans=bfs(z);
		printf("Case %d: %d\n",cas++,dis[ans]);	
	}
	return 0;
}