【BZOJ】1674: [Usaco2005]Part Acquisition(spfa)

http://www.lydsy.com/JudgeOnline/problem.php?id=1674

想法很简单。。。将每一种看做一个点,如果i可以换成j,那么连边到j。。

费用都为1.。

然后拥有过的物品就是最短路+1.。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1005, Q=N*100, oo=~0u>>2;
int q[Q], front, tail, d[N], vis[N], ihead[N], cnt, n, x;
struct ED { int to, next; }e[Q];
void add(int u, int v) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v;
}
int spfa() {
	q[tail++]=1;
	for1(i, 2, 1005) d[i]=oo;
	vis[1]=1;
	while(front!=tail) {
		int v, u=q[front++]; if(front==Q) front=0; vis[u]=0;
		for(int i=ihead[u]; i; i=e[i].next) if(d[v=e[i].to]>d[u]+1) {
			d[v]=d[u]+1;
			if(!vis[v]) {
				vis[v]=1;
				q[tail++]=v; if(tail==Q) tail=0;
			}
		}
	}
	return d[x];
}

int main() {
	read(n); read(x);
	for1(i, 1, n) {
		int u=getint(), v=getint();
		add(u, v);
	}
	int ans=spfa();
	if(ans==oo) puts("-1");
	else print(ans+1);
	return 0;
}

 

 


 

 

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

posted @ 2014-09-10 06:08  iwtwiioi  阅读(186)  评论(0编辑  收藏  举报