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

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=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';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,,sizeof(dis));

    dis[]=;

    q.push(make_pair(,));

    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=;i<=n;i++)

    {

        x=read();y=read();

        add(x,y,);

    }

    dijkstra();

    dis[k]>?puts("-1"):printf("%d\n",dis[k]+);

    return ;

}