poj1985 Cow Marathon（求树的直径）

Link：http://poj.org/problem?id=1985

Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 4351   Accepted: 2178 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 Hint 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.  Source USACO 2004 February

AC code：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<queue>
#define LL long long
using namespace std;
int n,m;
struct node{
    int from;
    int to;
    int w;
    int next;
}Edge[100011*4];
int head[100011];
int tot;
void init()
{
    memset(head,-1,sizeof(head));
    tot=0;
}
void add(int f,int t,int w)
{
    Edge[tot].from=f;
    Edge[tot].to=t;
    Edge[tot].w=w;
    Edge[tot].next=head[f];
    head[f]=tot++;
}
int dis[100011];
int vis[100011];
LL maxx;
int maxx_num;
void bfs(int st)
{
    queue<int>q;
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    int cur,nex;
    q.push(st);
    vis[st]=1;
    dis[st]=0;
    maxx=0;
    maxx_num=st;
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        for(int i=head[cur];i!=-1;i=Edge[i].next)
        {
            nex=Edge[i].to;
            if(!vis[nex])
            {
                vis[nex]=1;
                dis[nex]=dis[cur]+Edge[i].w;
                if(maxx<dis[nex])
                {
                    maxx=dis[nex];
                    maxx_num=nex;
                }
                q.push(nex);
            }
        }
    }
}
int main()
{
    int u,v,w;
    char s[5];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        while(m--)
        {
            scanf("%d%d%d%s",&u,&v,&w,&s);
            add(u,v,w);
            add(v,u,w);
        }
        bfs(1);
        bfs(maxx_num);
        printf("%lld\n",maxx);
    }
    return 0;
}