poj 2457(spfa+路径保存）

问题描述

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.

输入

* 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.

输出

* 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.

样例输入

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

样例输出

4
1
3
2
5      

套用SPFA模板+路径保存.

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

int pre[50001],dis[50001],vis[50001];int a[50001];int c[50001];
const int INF=99999999;
int n,k,top;

vector<int>vec[50001];

int spfa()
{
    queue<int> que;
    for(int i=1;i<=n;i++)
    {
        dis[i]=INF;
        vis[i]=0;
        pre[i]=1;
        c[i]=0;
    }
    que.push(1);
    vis[1]=1;dis[1]=0;
    while(!que.empty())
    {
        int x=que.front();
        que.pop();
        vis[x]=0;
        for(int i=0;i<vec[x].size();i++)
        {
            int y=vec[x][i];
            if(dis[x]+1<dis[y])
            {
                dis[y]=dis[x]+1;
                pre[y]=x;
                if(!vis[y])
                {
                    vis[y]=1;
                    c[y]++;
                    que.push(y);
                    if(c[y]>n)
                        return -1;
                }
            }
        }
    }
}

void add(int u,int v)
{
    vec[u].push_back(v);
}

int main()
{
    while(scanf("%d %d",&n,&k)!=EOF)
    {
         top=0;int L=k;
        for(int i=1;i<=n;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            add(a,b);
        }
        if(spfa()!=-1&&dis[k]!=INF)
        {
            int ans=dis[k]+1;
        printf("%d\n",ans);
        int kk=0;
        while(k!=1){
            a[++kk]=pre[k];
            k=pre[k];
        }
      
        while(kk>0)
            printf("%d\n",a[kk--]);
        printf("%d\n",L);
        }
        else
            printf("-1\n");
        
    }
}