LightOJ1094

http://lightoj.com/volume_showproblem.php?problem=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

树的直径典型题目 

树的直径是让求给你一棵树 所有节点和他们的距离然后求这棵树上距离最远的两个点之间的距离

分析：

先是任意找一个点，然后bfs找到离他最远距离的点，把这个点标记起来

然后用这个点再bfs一次找到直径

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <queue>

using namespace std;
#define N 30005
int ans,Max,Index;
int head[N],vis[N],dis[N];
struct node
{
    int v,f,next;
}e[N*4];

void Inn()
{
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    ans=Index=0;
}
void Add(int u,int v,int f)
{
    e[ans].v=v;
    e[ans].f=f;
    e[ans].next=head[u];
    head[u]=ans++;
}

void bfs(int u)
{
    Max=0;
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    queue<int>Q;
    Q.push(u);
    vis[u]=1;
    while(!Q.empty())
    {
        int p,q;
        p=Q.front();
        Q.pop();
        for(int i=head[p];i!=-1;i=e[i].next)
        {
            q=e[i].v;
            if(!vis[q])
            {
                vis[q]=1;
                Q.push(q);
                dis[q]=dis[p]+e[i].f;
                if(Max<dis[q])
                {
                    Max=dis[q];
                    Index=q;
                }
            }
        }
    }
}

int main()
{
    int T,n,U,V,W,t=1;
    scanf("%d",&T);
    while(T--)
    {
        Inn();
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d %d %d",&U,&V,&W);
            Add(U,V,W);
            Add(V,U,W);
        }
        bfs(0);
        bfs(Index);
        printf("Case %d: %d
",t++,Max);
    }
    return 0;
}