Til the Cows Come Home

又到了最短路径专题呢
先来几道基础题，回顾一下

题目描述：

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.

输入：

• 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.

输出：

• Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

样例输入：

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

样例输出：

90

发现写博客的时候，翻译还是很有必要的，机翻有时候实在是。。。。。一言难尽

题目大意：

有N个点，给出从a点到b点的距离，当然a和b是互相可以抵达的，问从1到n的最短距离
很明显，每条路都是双向联通的
我们可以采用dijkstra算法来实现
代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int inf=99999999;
int a[2005][2005];
int visit[2005];
int d[2005];
int n,m;
void djstera(int start)
{
	fill(d,d+n,inf);
	memset(visit,0,sizeof(visit));
	d[start]=0;
	while(1)
	{
		int v=-1;
		for(int i=0;i<n;i++)
		{
			if(!visit[i]&&(v==-1||d[i]<d[v]))
				v=i;
		}//找到未经搜索的那条最短路
		if(v==-1)break;
		visit[v]=1;
		for(int i=0;i<n;i++)
		{
			d[i]=min(d[i],d[v]+a[v][i]);
		}
	}
	printf("%d\n",d[n-1]);
}
int main()
{
	while(~scanf("%d%d",&m,&n))
	{
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(i!=j)a[i][j]=inf;
				else a[i][j]=0;
			}
		}
		int i,j;
		int x,y,d;
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d",&x,&y,&d);
			if(a[x-1][y-1]>d)
			a[x-1][y-1]=a[y-1][x-1]=d;//双向联通
		} 
		djstera(0);
	}
	return 0;
}