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
介个题就是求树的直径，但是自己还是不理解，没弄明白，代码不AC。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#define M 30000+10
using namespace std;
struct Edge{
int from,to,val,next;
};
Edge edge[M*4];
int head[M];//每一个节点的头位置
int edgenum;//总边数
int u,v,w,n;
int ans; //最后结果
int tnode; //记录S-T的端点
int dist[M];//以该点结尾的最长路
bool vis[M];//标记节点是否被访问
void init(){
memset(head,-1,sizeof(head));
tnode=0;edgenum=0;
}
void addeage(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;
edgenum++;
}
void BFS(int s)
{
ans=0;
memset(dist,0,sizeof(dist));
memset(vis,false,sizeof(vis));
queue<int > q;
vis[s]=true;dist[s]=0;ans=0;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.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;
}
vis[v]=true;
q.push(v);
}
}
}
for(int i=0;i<n;i++)
{
if(ans<dist[i])
{
ans=dist[i];
tnode=i;
}
}
}
int main()
{
int t,k=0;
scanf("%d",&t);
while(t--)
{
init();
scanf("%d",&n);
for(int i=1;i<n;i++)
{
scanf("%d%d%d",&u,&v,&w);
addeage(u,v,w);
addeage(v,u,w);
}
BFS(0);
BFS(tnode);
printf("Case %d: %d\n",++k,ans);
}
return 0;
}