poj 2631 树的直径裸题 注意设置n

题目：

Roads in the North

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2893 Accepted: 1414

Description

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice. 
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area. 

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1. 

Input

Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way.

Output

You are to output a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 61 4 56 3 92 6 86 1 7

Sample Output

22

代码：

#include <stdio.h>#include <vector>#include <queue>using namespace std;const int maxn=1e4+30;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){                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(){//588K32MS    int a,b,c;    for(int i=0;i<=maxn;++i) G[i].clear();    n=0;//要用到结点数 然而输入中没有显示给出.......    while(scanf("%d%d%d",&a,&b,&c)==3){        n=max(n,max(a,b));        G[a].push_back(edge(b,c));        G[b].push_back(edge(a,c));    }    printf("%d\n",solve());    return 0;}