[Usaco2005][BZOJ1674] Part Acquisition|dijkstra|priority_queue

1674: [Usaco2005]Part Acquisition

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 308  Solved: 143
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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

 

Source

Silver

 

 
抄了个堆优化dijkstra的模板……
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>
#define inf 1000000000
#define pa pair<int,int>
using namespace std;
int n,m,cnt,dis[1005],head[1005];
bool vis[1005];
int next[50005],list[50005],key[50005];
inline int read()
{
    int a=0,f=1; char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
    return a*f;
}
inline void insert(int u,int v,int w)
{
    next[++cnt]=head[u];
    head[u]=cnt;
    list[cnt]=v;
    key[cnt]=w;
}
inline void dijkstra()
{
    priority_queue<pa,vector<pa>,greater<pa> > q;
    for (int i=1;i<=n;i++) dis[i]=inf,vis[i]=0;
    dis[1]=0;
    q.push(make_pair(0,1));
    while (!q.empty())
    {
        int now=q.top().second; q.pop();
        if (vis[now]) continue; vis[now]=1;
        for (int i=head[now];i;i=next[i])
        {
            if (dis[list[i]]>dis[now]+key[i])
            {
                dis[list[i]]=dis[now]+key[i];
                q.push(make_pair(dis[list[i]],list[i]));
            }
        }
    }
}
int main()
{
    m=read(); n=read();
    for (int i=1;i<=m;i++) 
    {
        int u=read(),v=read();
        insert(u,v,1);
    }
    dijkstra();
    if (dis[n]==inf) puts("-1"); else printf("%d",dis[n]+1);
    return 0;
}

 

posted @ 2015-08-28 14:40  ws_fqk  阅读(266)  评论(0编辑  收藏  举报