POJ part acquisition

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.

 

寻找最短路径，输出路径

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

#define inf 0x3f3f3f3f

int n,k;
int dis[50001],w[1001][1001],pre[50002];
bool vis[50001];
int path[1001],ans[1001];

void dijkstra(int x)
{
    int i,j,Min,kk=0,next;
    for(i=1;i<=k;i++)
    {
        dis[i]=w[x][i];
        if(i!=x&&dis[i]<inf)
            path[i]=x;
    }
    vis[x]=1;
    dis[x]=0;
    for(i=1;i<k;i++)
    {
        Min=inf;
        for(j=1;j<=k;j++)
        {
            if(!vis[j]&&Min>dis[j])
            {
                Min=dis[j];
                next=j;
            }
        }
        vis[next]=1;
        for(j=1;j<=k;j++)
        {
            if(!vis[j]&&dis[next]+w[next][j]<dis[j]&&w[next][j]!=inf)
            {
                dis[j]=dis[next]+w[next][j];
                path[j]=next;//j下一个要放的点，next上一次循环中已经确定要放的点
                //cout<<j<<"  "<<path[j]<<endl;
            }
        }
    }
}

void print()
{
    int i;
    memset(ans,0,sizeof(ans));
    int cnt=0;
    if(dis[k]==inf)
        cout<<"-1"<<endl;
    else
    {
        ans[cnt++]=k;
        int temp=path[k];//下标是当前点，数组值为前一个节点
        while(temp)
        {
            //cout<<temp<<endl<<endl;
            ans[cnt++]=temp;
            temp=path[temp];
            //cout<<temp<<endl;
        }
        cout<<cnt<<endl;
        for(i=cnt-1;i>=0;i--)
            cout<<ans[i]<<endl;
    }
}

int main()
{
    int i,j,a,b;
    scanf("%d%d",&n,&k);
    memset(vis,0,sizeof(vis));
    memset(dis,inf,sizeof(dis));
    //memset(w,0x3f3f3f3f,sizeof(w));
    for(i=1;i<=k;i++)
        for(j=1;j<=k;j++)
            w[i][j]=inf;
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&a,&b);
        w[a][b]=1;
    }
    dijkstra(1);
    //cout<<dis[k]+1<<endl;
    print();
    return 0;
}