【light oj】树的直径

Farthest Nodes in a Tree

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

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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

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#include <cstdio>  
#include <cstring>  
#include <queue>   
#include <algorithm>   
using namespace std;  
struct Edge  
{  
    int from, to, val, next;  
};  
Edge edge[60000+10];  
int head[30000+10];//头'指针' 
int edgenum;//边数  
int dist[30000+10];  //以该点为尾点的最长距离 
bool vis[30000+10];  
int node; //记录端点值 
int ans;  //最终的最长路径 
int n;  
int k = 1;  
void init()  //初始化 
{  
    edgenum = 0;  
    memset(head, -1, sizeof(head));  
}  
void addedge(int u, int v, int w)  
{  
    edge[edgenum].from = u;//起点 
    edge[edgenum].to = v;//终点 
    edge[edgenum].val = w;//权值 
    edge[edgenum].next = head[u];//指向下一条边 
    head[u] = edgenum++;  
}  
void BFS(int s)  
{  
    queue<int> q;  
    memset(dist, 0, sizeof(dist));  
    memset(vis, false, sizeof(vis));  
    vis[s] = true;  
    ans = 0;  
    node = s;  
    q.push(s);  
    while(!q.empty())  
    {  
        int u = q.front();  
       q.pop();  
        for(int i = head[u]; i != -1; i = edge[i].next)  //遍历每一条边 
        {  
           int v = edge[i].to;  
            if(!vis[v] && dist[v] < dist[u] + edge[i].val)  
            {  
                vis[v] = true;  
                dist[v] = dist[u] + edge[i].val;  
                if(dist[v] > ans)  
                {  
                    ans = dist[v];  
                    node = v;  
                }  
               q.push(v);  
            }  
        }  
    }  
}   
int main()  
{  
    int t;  
    scanf("%d", &t);  
    while(t--)  
    {  
        scanf("%d", &n);  
        init();  
         int a, b, c;  
    for(int i = 1; i < n; i++)  
    {  
        scanf("%d%d%d", &a, &b, &c);  
        //a++, b++;  
        addedge(a,b,c);  
        addedge(b,a,c);  
    }   
         BFS(1);  //①任意从一个点u出发bfs，设其能到的最远点为v,0不可以 
         BFS(node);  //②从v出发重新bfs，设其能到达的最远点为s
    printf("Case %d: %d\n", k++, ans);  
  
    }  
    return 0;  
}