Cow Marathon（树的直径）（最长路）

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. 

* Lines 1.....: Same input format as "Navigation Nightmare".

* Line 1: An integer giving the distance between the farthest pair of farms. 

The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52. 

套模板：

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 300010;
struct Edge{
	int from,to,val,next;
};
Edge edge[2*MAXN];

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 n;
int ans;
int longest;
int dis[MAXN];
bool vis[MAXN];
void BFS(int x)
{
	memset(dis,0,sizeof(dis));
	memset(vis,false,sizeof(vis));
	queue<int> Q;
	Q.push(x);
	dis[x]=0;
	vis[x]=true;
	ans=0;
	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(dis[v]<dis[u]+edge[i].val){
					dis[v]=dis[u]+edge[i].val;
					/*if(ans<dis[v]){  把这个放在下面的for循环中会快一些；
						ans=dis[v];
						longest=v;
					}*/
				}
				vis[v]=true;
				Q.push(v); 
			}
		}
	}
	for(int i = 1; i <= n; i++) {	//这里要注意，有的图的顶点不是从 0 开始的；比如这次的农场，没有第零个农场吧；在主函数中BFS时也要注意； 
        if(ans < dis[i]) {
          	ans = dis[i];
          	longest = i;
        }
    }
}

int main()
{
	char s[2];
	int m,a,b,c;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d%s",&a,&b,&c,s);
			getchar();
			addEdge(a,b,c);
			addEdge(b,a,c);
		}BFS(1);  BFS(longest);		//BFS(1);中的1是任意找的，也可以是2...但不能是零； 
		printf("%d\n",ans);
	}
	return 0;
}

</pre><pre name="code" class="cpp">

再帮助大家理解理解：

<pre name="code" class="cpp">#include<iostream> 
#include<queue>
#define Maxn 200 
using namespace std; 
struct Edge{
	int from,to,val,next;
}edge[Maxn];	//存储边信息的结构体 
 
int head[Maxn];		//起点为下标存储(edge中边的位置)  
int main() 
{ 

     int edgenum;	//边数  
     memset(head,-1,sizeof(head)); 
     //因为刚开始head不指向任何一条边的下标,所以head都为-1  
     cin>>edgenum;		//边数  
     for(int i=0;i<edgenum;i++) 
     { 
         cin>>edge[i].from>>edge[i].to>>edge[i].val;	//起点 终点 权重  
         edge[i].next=head[edge[i].from];
		 head[edge[i].from]=i;		//不容易理解的地方 
         /*
         利用head数组存储的是最新的(以数组下标为起点的)边的下标 
         并且该条边的next指向的是同样以数组下标为起点的下一条边的下标 
         直到下一条边的next=-1(即将所有以数组下标为起点的边都遍历了一遍) 
         */ 
     } 
     for(int u=1;u<10;u++)	//输出图  
     { 
        cout<<"以"<<u<<"为起点的所有边的信息："<<endl;  
        for(int v=head[u];v!=-1;v=edge[v].next)		//遍历以u为起点的所有边的信息  
        cout<<edge[v].from<<" "<<edge[v].to<<" "<<edge[v].val<<endl;
     } 
     return 0; 
}