# BZOJ1674: [Usaco2005]Part Acquisition

## 1674: [Usaco2005]Part Acquisition

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 259  Solved: 114
[Submit][Status]

## 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.

## Source  1 #include<cstdio>
2 #include<cstdlib>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<iostream>
7 #include<vector>
8 #include<map>
9 #include<set>
10 #include<queue>
11 #define inf 1000000000
12 #define maxn 50000+100
13 #define maxm 2000000
14 #define eps 1e-10
15 #define ll long long
16 #define pa pair<int,int>
17 using namespace std;
18 inline int read()
19 {
20     int x=0,f=1;char ch=getchar();
21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
22     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
23     return x*f;
24 }
25 int n,m,tot;
26 int d,head;
27 bool v;
28 struct edge{int go,next,w;}e;
29 void insert(int x,int y,int z)
30 {
31     e[++tot].go=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
32 }
33 void dijkstra()
34 {
35     priority_queue<pa,vector<pa>,greater<pa> >q;
36     for(int i=1;i<=n;i++)d[i]=inf;
37     memset(v,0,sizeof(v));
38     d=0;q.push(make_pair(0,1));
39     while(!q.empty())
40     {
41         int x=q.top().second;q.pop();
42         if(v[x])continue;v[x]=1;
43         for(int i=head[x],y;i;i=e[i].next)
44             if(d[x]+e[i].w<d[y=e[i].go])
45             {
46                 d[y]=d[x]+e[i].w;
47                 q.push(make_pair(d[y],y));
48             }
49
50     }
51 }
52 int main()
53 {
54     freopen("input.txt","r",stdin);
55     freopen("output.txt","w",stdout);
56     m=read();n=read();
57     for(int i=1;i<=m;i++)
58     {
59         int x=read(),y=read(),z=1;
60         insert(x,y,z);
61     }
62     dijkstra();
63     if(d[n]==inf)puts("-1");
64     else printf("%d",d[n]+1);
65     return 0;
66 }
View Code

posted @ 2014-08-21 17:00  ZYF-ZYF  Views(169)  Comments(0Edit  收藏  举报