# 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.

## 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

#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>
{
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()
{