Farthest Nodes in a Tree

最新推荐文章于 2019-06-28 18:16:29 发布

原创最新推荐文章于 2019-06-28 18:16:29 发布 · 237 阅读

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本文介绍了一种算法，用于在加权无向树中找到距离最远的两个节点。输入包含多个测试用例，每个用例由节点数及连接各节点的边组成，输出为最远两点间的最大距离。

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

0 1 20

1 2 30

2 3 50

0 2 20

2 1 10

0 3 29

0 4 50

Sample Output

Case 1: 100

Case 2: 80

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define M 31000
struct node {
    int v,next,w;
}mp[M*3];
int cnt,head[M],dis[M],vis[M];
int ans,last;
void add(int u,int v,int w){
    mp[cnt].v=v;
    mp[cnt].w=w;
    mp[cnt].next=head[u];
    head[u]=cnt++;
}
void bfs(int s){
    queue<int> q;
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    vis[s]=1;
    last=s;
    ans=0;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=mp[i].next){
            int v=mp[i].v;
            if(!vis[v] && dis[v]<dis[u]+mp[i].w){
                vis[v]=1;
                dis[v]=dis[u]+mp[i].w;
                if(ans<dis[v]){
                    ans=dis[v];
                    last=v;  
                }  
                q.push(v);  
            }  
        }  
    }  
}  
int main(){  
    int t,n,a,b,c,k=1;  
    scanf("%d",&t);  
    while(t--){  
        scanf("%d",&n);  
        cnt=0;  
        memset(head,-1,sizeof(head));  
        for(int i=0;i<n-1;i++){  
            scanf("%d %d %d",&a,&b,&c);  
            add(a,b,c);  
            add(b,a,c);  
        }  
        bfs(0);  
        bfs(last);  
        printf("Case %d: ", k++);   
        printf("%d\n",ans);   
    }  
    return 0;  
}