最短路径的补充——— Til the Cows Come Home

https://vjudge.net/problem/POJ-2387
原文：
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
  题意描述：
  找出n到1号的最短距离
  解题思路：
  用单源最短路径去求，求出后输出dis[1]就行，写的话很容易，就是在输入的时候容易有陷阱，例如给两组数：
  2 1 3
  2 1 6
  这样的话本来2->1的最短路是3结果被后来输入的6覆盖了；
  AC代码：

#include<stdio.h>
#include<string.h>
int inf=99999999;
int book[10050];
int e[10000][10000],dis[10500];
int main()
{
    int n,m,i,j,min,u,v,w;
    while(scanf("%d %d", &m,&n)!=EOF)
    {
    //初始化数组
    	for(i=1; i<=n; i++)
       {
       	for(j=1; j<=n; j++)
       	{
       		if(i==j)
       		e[i][j]=0;
       		else
       		e[i][j]=inf;
   		}
   	}
   	//输入数据
      for(i=1; i<=m; i++)
       {
       	scanf("%d %d %d",&u,&v,&w);
       	{
       	//题目所设陷阱的地方
       		if(w<e[u][v])
       		{ 
       			e[u][v]=e[v][u]=w;
       		}
   		}
       
   	}
               for(i=1; i<=n; i++)
   			{
               	dis[i]=e[1][i];
   			}
   			memset(book,0,sizeof(book));
   			   book[1]=1;
   			 for(i=1; i<=n; i++)
   			 {
   			     min=inf;	
   			 	for(j=1; j<=n; j++)
   			 	{
   			 	 if(book[j]==0&&min>dis[j])
   				{
   					 u=j;
   					 min=dis[j];
   					}
   				 }
   				 book[u]=1;
   				 for(v=1; v<=n; v++)
   				 {
   				 	if(e[u][v]<inf)
   				 	{
   				 		if(dis[v]>e[u][v]+dis[u])
   				 		  dis[v]=e[u][v]+dis[u];
   					 }
   				 }
   			  } 
         printf("%d\n", dis[n]);
    }  
   return 0;
}