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.

HINT

 

Source

Silver

题解：

我会说我没看题就是为了写个dijkstra+heap的模版吗？

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #define inf 1000000000
 #define maxn 50000+100
 #define maxm 2000000
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 using namespace std;
 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 n,m,tot;
 int d[],head[];
 bool v[];
 struct edge{int go,next,w;}e[];
 void insert(int x,int y,int z)
 {
     e[++tot].go=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
 }
 void dijkstra()
 {
     priority_queue<pa,vector<pa>,greater<pa> >q;
     for(int i=;i<=n;i++)d[i]=inf;
     memset(v,,sizeof(v));
     d[]=;q.push(make_pair(,));
     while(!q.empty())
     {
         int x=q.top().second;q.pop();
         if(v[x])continue;v[x]=;
         for(int i=head[x],y;i;i=e[i].next)
             if(d[x]+e[i].w<d[y=e[i].go])
             {
                 d[y]=d[x]+e[i].w;
                 q.push(make_pair(d[y],y));
             }

     }
 }
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     m=read();n=read();
     for(int i=;i<=m;i++)
     {
         int x=read(),y=read(),z=;
         insert(x,y,z);
     }
     dijkstra();
     if(d[n]==inf)puts("-1");
     else printf("%d",d[n]+);
     return ;
 }