poj 1985 Cow Marathon【树的直径裸题】

Cow Marathon

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 4185		Accepted: 2118
Case Time Limit: 1000MS

Description

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. 

Input

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

Output

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

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S

Sample Output

52
昨天比赛遇到一个树的直径的题 不知道是什么  唉  坑啊，今天赶紧看看
题意：N个点M条边，每条边都有权值，让你找到两点，使得这两点间所有边的权值之和最长，
     （这条路径上任意节点不能有岔路（此处不能有岔路指的是岔路上的边的权值不能算上））
题解：运用两次bfs求最长距离 第一次先任选一点，找到这一点可以到达的最长距离的节点u处
     然后再以u点为起点，找到最长路

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 100000
using namespace std;
int head[MAX];
int vis[MAX],dis[MAX];
int n,m,ans;
int sum,aga;
char s[30];
struct node
{
	int u,v,w;
	int next;
}edge[MAX];

void add(int u,int v,int w)
{
	edge[ans].u=u;
	edge[ans].v=v;
	edge[ans].w=w;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void getmap()
{
	int i,j;
	int a,b,c;
	ans=0;
	memset(head,-1,sizeof(head));
    while(m--)
    {
    	scanf("%d%d%d%s",&a,&b,&c,s);
    	add(a,b,c);
    	add(b,a,c);
	}
}

void bfs(int beg)
{
	queue<int>q;
	memset(dis,0,sizeof(dis));
	memset(vis,0,sizeof(vis));
	int i,j;
	while(!q.empty())
	    q.pop();
	aga=beg;
	sum=0;
	vis[beg]=1;
	q.push(beg);
	int top;
	while(!q.empty())
	{
		top=q.front();
		q.pop();
		for(i=head[top];i!=-1;i=edge[i].next)
		{
			if(!vis[edge[i].v])
			{
				dis[edge[i].v]=dis[top]+edge[i].w;
				vis[edge[i].v]=1;
				q.push(edge[i].v);
			}
		}
	}

	for(i=1;i<=n;i++)
	{
		if(sum<dis[i])
		{
			sum=dis[i];
			aga=i;
		}
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		getmap();
		bfs(1);
		bfs(aga);
		printf("%d\n",sum);
	}
	return 0;
}