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=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
return x*f;
}
int n,m,tot;
int d[1005],head[1005];
bool v[1005];
struct edge{int go,next,w;}e[50005];
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=1;i<=n;i++)d[i]=inf;
memset(v,0,sizeof(v));
d[1]=0;q.push(make_pair(0,1));
while(!q.empty())
{
int x=q.top().second;q.pop();
if(v[x])continue;v[x]=1;
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=1;i<=m;i++)
{
int x=read(),y=read(),z=1;
insert(x,y,z);
}
dijkstra();
if(d
==inf)puts("-1");
else printf("%d",d
+1);
return 0;
}

View Code