1094

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1094 - Farthest Nodes in a Tree

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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-1lines 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

2

4

0 1 20

1 2 30

2 3 50

5

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.

思路：
树的直径的裸题，用前向星存图（多用），从任意一点出发广搜，bfs出一个最长路径的叶子节点id（这个节点就是树的直径的一个端点），再从id点出发二次深搜，这次搜出最长路径的叶子节点就是树的直径的另外一个节点，那么第二次深搜的这个路径就是树的直径，那么这样两次深搜就可以找出树的直径；

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<cmath>#include<vector> #define max_n 100010using  namespace std;typedef long long LL;int head[max_n],dis[max_n],vis[max_n];int t,n,ans,id=0;struct node{int to;int val;int next;}edge[max_n];void addedge(int u,int v,int w){ //添加边，前向星存图 node e={v,w,head[u]};  edge[id]=e;head[u]=id++;}void bfs(int x){queue<int> q;memset(dis,0,sizeof(dis));memset(vis,0,sizeof(vis));vis[x]=1;ans=0,id=x;q.push(x);while(!q.empty()){int p=q.front();q.pop();for(int i=head[p];i!=-1;i=edge[i].next){if(!vis[edge[i].to] && dis[edge[i].to]<dis[p]+edge[i].val){vis[edge[i].to]=1;dis[edge[i].to]=dis[p]+edge[i].val; //更新点 if(dis[edge[i].to]>ans){ans=dis[edge[i].to]; //比较出最长的路径，记录下节点 id=edge[i].to;}q.push(edge[i].to);}}}}int main(){int a,b,c;scanf("%d",&t);while(t--){ans=0;memset(head,-1,sizeof(head));scanf("%d",&n);for(int i=0;i<n-1;i++){scanf("%d %d %d",&a,&b,&c);a++,b++; addedge(a,b,c); //无向图 addedge(b,a,c);}bfs(1); //任意一点bfs(id); //直径的一个端点 static int p=1;printf("Case %d: %d\n",p++,ans);}return 0;}