lightoj 1094

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

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

要求树上距离最远的两点之间的距离，也就是要求树的直径，树的直径上的两个端点肯定是过根结点的，所以可以先搜出一个端点，再从这个端点把另一个端点搜出来，求出的距离就是树的直径。

#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct edge{    int to,cost;};vector<edge> e[60010];int farthest,ans;void dfs(int x,int pre,int dis){    for(int i=0;i<e[x].size();i++)    {        int xx = e[x][i].to;        if(xx == pre)            continue;        dfs(xx,x,dis+e[x][i].cost);    }    if(dis > ans)    {        ans = dis;        farthest = x;    }}int main(void){    int T,n,i,j;    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d",&n);        for(i=0;i<=n;i++)            e[i].clear();        for(i=0;i<n-1;i++)        {            int x,y;            edge t;            scanf("%d%d%d",&x,&y,&t.cost);            t.to = y;            e[x].push_back(t);            t.to = x;            e[y].push_back(t);        }        ans = 0;        dfs(0,-1,0);        dfs(farthest,-1,0);        printf("Case %d: %d\n",cas++,ans);    }    return 0;}