poj 1985 树的直径模板题

题目：

Cow Marathon

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 5325 Accepted: 2614Case 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 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S

n个点,m条边,每条边给出两个端点和距离，和方向（即东西南北，本题中该信息无用） 求图中两点之间的最大距离

挑战2代码：

#include <stdio.h>#include <vector>#include <queue>using namespace std;const int maxn=1e5+20;const int inf=0x3f3f3f3f;struct edge{    int t,w;    edge(int tt=0,int ww=0):t(tt),w(ww){}};vector<edge> G[maxn];int n,m,d[maxn];//bool vis[maxn];//int cnt;void bfs(int s){    for(int i=0;i<=n;++i) d[i]=inf;    queue<int> q;    q.push(s);    d[s]=0;    int u;    while(!q.empty()){        u=q.front(); q.pop();        for(int i=0;i<G[u].size();++i){            edge e=G[u][i];            if(d[e.t]==inf){// or d[e.t]>d[u]+e.w 一样的 因为从树上一点出发不可能通过两个不同的点到达一个点 即无论如何只会更新一次，入队一次                d[e.t]=d[u]+e.w;                q.push(e.t);            }        }    }}int solve(){    bfs(1);//任选一个节点出发 如果下标是从1开始的不能bfs(0).....!!!    int maxv=0;    int maxi=0;    for(int i=0;i<=n;++i){        if(d[i]==inf) continue;        if(maxv<d[i]){            maxv=d[i];            maxi=i;        }    }    bfs(maxi);    maxv=0;    for(int i=0;i<=n;++i){        if(d[i]==inf) continue;        maxv=max(maxv,d[i]);    }    return maxv;}int main(){//3928K1141MS    int a,b,c;    char s[3];    while(scanf("%d%d",&n,&m)==2){        for(int i=0;i<=n;++i) G[i].clear();        while(m--){            scanf("%d%d%d%s",&a,&b,&c,s);            //cout<<a<<" "<<b<<" "<<c<<" "<<s<<endl;            G[a].push_back(edge(b,c));            G[b].push_back(edge(a,c));        }        printf("%d\n",solve());    }    return 0;}