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

    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
    ",k++,ans);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273514.html
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