POJ 2457 Part Acquisition

Part Acquisition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2909		Accepted: 1262		Special Judge

Description

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.

Source

USACO 2005 February Silver

 

 

题意：cows 想用自己手上的商品（1）通过多次交换得到想要的商品（k），给出两种商品的交换关系，求出至少有多少种商品经过交换，输出交换的顺序。

思路：最短路。相当于求从1商品到k商品的最短路径，中间再记录一下结点信息。我用的是边表实现dij算法

 

  1 #include<iostream>
  2 #include<queue>
  3 #include<cstdio>
  4 #include<cstring>
  5 
  6 using namespace std;
  7 
  8 const int maxn=50010;
  9 const int INF=0x3f3f3f3f;
 10 
 11 struct PP{
 12     int id;
 13     int len;
 14     bool operator < (const PP &a) const{
 15         return a.len<len;
 16     }
 17 }dis[maxn];
 18 
 19 int vis[maxn],head[maxn],pre[maxn],ans[maxn];
 20 int id;
 21 
 22 struct node{
 23     int v,w;
 24     int next;
 25 }e[100010];
 26 
 27 void addedge(int u,int v,int w){    //有向边 
 28     e[id].v=v;
 29     e[id].w=w;
 30     e[id].next=head[u];
 31     head[u]=id;
 32     id++;
 33 }
 34 
 35 int n,src,des;
 36 
 37 void Dijkstra(){
 38     priority_queue<PP> Q;
 39     while(!Q.empty())
 40         Q.pop();
 41     memset(vis,0,sizeof(vis));
 42     for(int i=1;i<=n;i++){
 43         dis[i].id=i;
 44         dis[i].len=INF;
 45     }
 46     dis[src].len=0;
 47     dis[src].id=src;
 48     Q.push(dis[src]);
 49     PP tmp;
 50     while(!Q.empty()){
 51         tmp=Q.top();
 52         Q.pop();
 53         int now=tmp.id;
 54         if(vis[now])
 55             continue;
 56         vis[now]=1;
 57         for(int i=head[now];i!=-1;i=e[i].next){
 58             int len=e[i].w;
 59             int to=e[i].v;
 60             if(dis[to].len>dis[now].len+len){
 61                 dis[to].len=dis[now].len+len;
 62                 pre[to]=now;
 63                 Q.push(dis[to]);
 64             }
 65         }
 66     }
 67 }
 68 
 69 int main(){
 70 
 71     //freopen("input.txt","r",stdin);
 72 
 73     src=1;
 74     while(~scanf("%d%d",&n,&des)){
 75         id=0;
 76         memset(head,-1,sizeof(head));
 77         int u,v;
 78         for(int i=1;i<=n;i++){
 79             scanf("%d%d",&u,&v);
 80             addedge(u,v,1);
 81         }
 82         Dijkstra();
 83         if(dis[des].len!=INF){
 84             printf("%d\n",dis[des].len+1);
 85             int cnt=0;
 86             ans[cnt++]=des;
 87             int cur=des;
 88             while(pre[cur]!=1){
 89                 ans[cnt++]=pre[cur];
 90                 cur=pre[cur];
 91             }
 92             ans[cnt]=1;
 93             for(int i=cnt;i>=0;i--){
 94                 printf("%d",ans[i]);
 95                 if(i!=0)
 96                     printf("\n");
 97             }
 98         }else
 99             printf("-1\n");
100     }
101     return 0;
102 }