Part Acquisition--（最短路径，Dijkstra）

The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying through a cluster of stars containing N (1 <= N <= 50,000) planets, each with a trading post. 

The cows have determined which of K (1 <= K <= 1,000) types of objects (numbered 1..K) each planet in the cluster desires, and which products they have to trade. No planet has developed currency, so they work under the barter system: all trades consist of each
party trading exactly one object (presumably of different types). 

The cows start from Earth with a canister of high quality hay (item 1), and they desire a new milking machine (item K). Help them find the best way to make a series of trades at the planets in the cluster to get item K. If this task is impossible, output -1.

Input
* Line 1: Two space-separated integers, N and K. 

* Lines 2..N+1: Line i+1 contains two space-separated integers, a_i and b_i respectively, that are planet i's trading trading products. The planet will give item b_i in order to receive item a_i.

Output
* Line 1: One more than the minimum number of trades to get the milking machine which is item K (or -1 if the cows cannot obtain item K). 

* Lines 2..T+1: The ordered list of the objects that the cows possess in the sequence of trades.

Sample Input

6 5
1 3
3 2
2 3
3 1
2 5
5 4

Sample Output

4
1
3
2
5

Hint
OUTPUT DETAILS: 

The cows possess 4 objects in total: first they trade object 1 for object 3, then object 3 for object 2, then object 2 for object 5.

题意：有n个交换点，k种物品，每个交换点对应a，b两个数表示可在此交换点用a换到b，问用1物品最少几次可以换到k物品，并输出路径

先想到是最短路径问题，用Dijktra来求，并最终判断是否能换到

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;
#define pi acos(-1.0)
#define inf 0x3f3f3f
#define M 1005
int n,k,s,e;
int a[50001][1001],num[M],dis[M],pre[M]; //a数组存放点的关系，num[i]是以第i个点为起点的边的个数，
bool ok[M]; //dis[i]是点i到1点的最少步数，pre[i]存放点的前驱节点
void print(int i)
{
if(pre[i]!=0) print(pre[i]);
printf("%d\n",i);
}
int main()
{
int i,j,t;
scanf("%d%d",&n,&k);
for(i=1;i<=n;i++)
{
scanf("%d%d",&s,&e);
a[s][num[s]++]=e; //点s的第num[s]条边的终点是e
}
memset(dis,127/3,sizeof(dis));
memset(ok,false,sizeof(ok));
dis[1]=1; pre[1]=0;
for(i=1;i<=k-1;i++)
{
t=0;
for(j=1;j<=k;j++)
if(!ok[j]&&dis[j]<dis[t])
t=j;
//cout<<t<<endl;
if(t==0) break;
ok[t]=true;
for(j=0;j<num[t];j++)
{
if(dis[t]+1<dis[a[t][j]])
{
dis[a[t][j]]=dis[t]+1;
pre[a[t][j]]=t;
}
}
}
if(dis[k]<k*2) {printf("%d\n",dis[k]); print(k);}
else {printf("-1\n");return 0;}
return 0;
}