【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;

}

 

 

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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