Til the Cows Come Home（最短路（单源SPFA））

Link：http://poj.org/problem?id=2387

Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31889   Accepted: 10809 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.  Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.  Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists. Input * Line 1: Two integers: T and N  * Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100. Output * Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1. Sample Input 5 5 1 2 20 2 3 30 3 4 20 4 5 20 1 5 100 Sample Output 90 Hint INPUT DETAILS:  There are five landmarks.  OUTPUT DETAILS:  Bessie can get home by following trails 4, 3, 2, and 1. Source USACO 2004 November

AC code：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define MAXN 20020
#define INF 9999999999
using namespace std;
int n,m;
LL d[MAXN];
struct node{
	int u,v,w;
	int next;
}E[MAXN];
bool vis[MAXN];
int head[MAXN];
int cnt[MAXN];
int tot;
void init()
{
	memset(head,-1,sizeof(head));
	tot=0;
}
void add(int uu,int vv,int ww)
{
	E[tot].u=uu;
	E[tot].v=vv;
	E[tot].w=ww;
	E[tot].next=head[uu];
	head[uu]=tot++;
}
bool spfa(int s,int n)
{
	queue<int>q;
	for(int i=0;i<=n;i++)
	{
		vis[i]=false;
		cnt[i]=0;
		d[i]=INF;
	}
	q.push(s);
	cnt[s]++;
	vis[s]=true;
	d[s]=0;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=false;
		if(cnt[u]>n)
			return false;
		for(int i=head[u];i!=-1;i=E[i].next)
		{
			int v=E[i].v;
			if(d[v]>E[i].w+d[u])
			{
				d[v]=d[u]+E[i].w;
				if(!vis[v])
				{
					q.push(v);
					vis[v]=true;
					cnt[v]++;
				}
			}
		}
	}
	return true;
}
int main()
{
	int u,v,w;
	while(cin>>m>>n)
	{
		init();
		while(m--)
		{
			cin>>u>>v>>w;
			add(u,v,w);
			add(v,u,w);//注意：因为是无向图，要双向加边 ！！！ 
		}
		spfa(n,n);//源点为n，终端为1 
		cout<<d[1]<<endl;
	}
	return 0;
}