poj 1985 Cow Marathon 树的直径

题目链接：http://poj.org/problem?id=1985

After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms.

题意描述：题意很简单，就是求解两个农场之间最远的距离是多少。

算法分析：树的直径。详细分析：hdu4607

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define inf 0x7fffffff
 using namespace std;
 const int maxn=+;
 const int M = +;

 int n,m;
 struct node
 {
     int v,w;
     int next;
 }edge[M*];
 int head[maxn],edgenum;

 void add(int u,int v,int w)
 {
     edge[edgenum].v=v ;edge[edgenum].w=w ;
     edge[edgenum].next=head[u] ;head[u]=edgenum++ ;

     edge[edgenum].v=u ;edge[edgenum].w=w ;
     edge[edgenum].next=head[v] ;head[v]=edgenum++ ;
 }

 int d[maxn],vis[maxn];
 int bfs(int from)
 {
     memset(d,-,sizeof(d));
     memset(vis,,sizeof(vis));
     queue<int> Q;
     Q.push(from);
     vis[from]= ;d[from]= ;
     int maxlen=-,k=;
     while (!Q.empty())
     {
         int u=Q.front() ;Q.pop() ;
         for (int i=head[u] ;i!=- ;i=edge[i].next)
         {
             int v=edge[i].v;
             if (!vis[v])
             {
                 vis[v]=;
                 d[v]=d[u]+edge[i].w;
                 Q.push(v);
                 if (d[v]>maxlen)
                 {
                     maxlen=d[v];
                     k=v;
                 }
             }
         }
     }
     return k;
 }

 int main()
 {
     while (scanf("%d%d",&n,&m)!=EOF)
     {
         memset(head,-,sizeof(head));
         edgenum=;
         int a,b,c;
         char str[];
         for (int i= ;i<=m ;i++)
         {
             scanf("%d%d%d%s",&a,&b,&c,str);
             add(a,b,c);
         }
         int v=bfs();
         int u=bfs(v);
         printf("%d\n",d[u]);
     }
     return ;
 }