bzoj 1674: [Usaco2005]Part Acquisition -- dijkstra（堆优化）

1674: [Usaco2005]Part Acquisition

Time Limit: 5 Sec  Memory Limit: 64 MB

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.

 

这些牛被派去太空任务，为他们的谷仓购置一台新挤奶机。他们是通过一组包含N（1 < = N = 50000）行星，每一个交易岗位的恒星飞行。奶牛已经确定的K（1 < = K = 1000）类型的对象（编号为1 K）每个行星在集群的欲望，以及他们必须交易的产品。没有行星发展货币，所以他们在易货制度下工作：所有交易由每一方交易完全一个对象（大概是不同类型）。奶牛从地球开始与高品质的干草筒（项目1），他们希望一个新的挤奶机（项目k）。帮助他们找到最好的方法，使一系列的交易在行星的集群，以获得项目K.如果这个任务是不可能的，输出- 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).

Sample Input

6 5 //6个星球,希望得到5,开始时你手中有1号货物.
1 3 //1号星球,希望得到1号货物,将给你3号货物
3 2
2 3
3 1
2 5
5 4

Sample Output

4


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.

HINT

题目就是相当于有n条有向边，边权为1，求1到k的最短路

#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define M 50010
#define N 1010
#define pa pair<int,int>
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int cnt,lj[N],fro[M],to[M],v[M];
void add(int a,int b,int w){fro[++cnt]=lj[a];to[cnt]=b;v[cnt]=w;lj[a]=cnt;}
int dis[N];
bool vs[N];
void dijkstra()
{
    priority_queue<pa,vector<pa>,greater<pa> >q;
    memset(dis,127,sizeof(dis));
    dis[1]=0;
    q.push(make_pair(0,1));
    int u;
    while(!q.empty())
    {
        u=q.top().second;q.pop();
        for(int i=lj[u];i;i=fro[i])
        {
            if(dis[to[i]]>dis[u]+v[i])
            {
                dis[to[i]]=dis[u]+v[i];
                q.push(make_pair(dis[to[i]],to[i]));
            } 
        }
    }
}
int n,k,x,y;
int main()
{
    n=read();k=read();
    for(int i=1;i<=n;i++)
    {
        x=read();y=read();
        add(x,y,1);
    }
    dijkstra();
    dis[k]>100000000?puts("-1"):printf("%d
",dis[k]+1);
    return 0;
}