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

 

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn=50010;
const int INF=0x3f3f3f3f;

struct PP{
    int id;
    int len;
    bool operator < (const PP &a) const{
        return a.len<len;
    }
}dis[maxn];

int vis[maxn],head[maxn],pre[maxn],ans[maxn];
int id;

struct node{
    int v,w;
    int next;
}e[100010];

void addedge(int u,int v,int w){    //有向边 
    e[id].v=v;
    e[id].w=w;
    e[id].next=head[u];
    head[u]=id;
    id++;
}

int n,src,des;

void Dijkstra(){
    priority_queue<PP> Q;
    while(!Q.empty())
        Q.pop();
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        dis[i].id=i;
        dis[i].len=INF;
    }
    dis[src].len=0;
    dis[src].id=src;
    Q.push(dis[src]);
    PP tmp;
    while(!Q.empty()){
        tmp=Q.top();
        Q.pop();
        int now=tmp.id;
        if(vis[now])
            continue;
        vis[now]=1;
        for(int i=head[now];i!=-1;i=e[i].next){
            int len=e[i].w;
            int to=e[i].v;
            if(dis[to].len>dis[now].len+len){
                dis[to].len=dis[now].len+len;
                pre[to]=now;
                Q.push(dis[to]);
            }
        }
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    src=1;
    while(~scanf("%d%d",&n,&des)){
        id=0;
        memset(head,-1,sizeof(head));
        int u,v;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&u,&v);
            addedge(u,v,1);
        }
        Dijkstra();
        if(dis[des].len!=INF){
            printf("%d\n",dis[des].len+1);
            int cnt=0;
            ans[cnt++]=des;
            int cur=des;
            while(pre[cur]!=1){
                ans[cnt++]=pre[cur];
                cur=pre[cur];
            }
            ans[cnt]=1;
            for(int i=cnt;i>=0;i--){
                printf("%d",ans[i]);
                if(i!=0)
                    printf("\n");
            }
        }else
            printf("-1\n");
    }
    return 0;
}