Light OJ 1049 Farthest Nodes in a Tree(树的直径)(模板题)

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

题意：给出n个结点，和n-1条边，求最长的一条边，即树的直径.

树的定义：没有回路的无向连通图。

树的性质：

1,一棵树中的任意两个结点有且仅有唯一的一条路径连通.

2,一棵树如果有n个结点，那么它一定恰好有n-1条边.

3,在一棵树中加一条边会构成一个回路.

代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN=30000+10;
struct node{
int from,to,next,val;
}edge[MAXN*2];
int head[MAXN];
int edgenum;
void init(){
memset(head,-1,sizeof(head));
edgenum=0;
}
void addedge(int u,int v,int w){
edge[edgenum].from=u;
edge[edgenum].to=v;
edge[edgenum].val=w;
edge[edgenum].next=head[u];
head[u]=edgenum++;
}
int ans=0;
int pt;
int dist[MAXN];
bool vis[MAXN];
int n;
void bfs(int s)
{
memset(dist,0,sizeof(dist));
memset(vis,0,sizeof(vis));
queue<int>que;
que.push(s);
vis[s]=true;
dist[s]=0;
ans=0;
pt=s;
while(!que.empty())
{
int u=que.front();
que.pop();
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(!vis[v])
{
if(dist[v]<dist[u]+edge[i].val)
{
dist[v]=dist[u]+edge[i].val;
if(ans<dist[v])
{
ans=dist[v];
pt=v;
}

}

vis[v]=true;
que.push(v);
}

}
}

}
int main()
{
int t,a,b,c,k=0;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d",&n);

for(int i=0;i<n-1;i++)
{
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);
addedge(b,a,c);
}
bfs(0);
bfs(pt);
printf("Case %d: %d\n",++k,ans);
}
}