bzoj 1674: [Usaco2005]Part Acquisition（最短路）

1674: [Usaco2005]Part Acquisition

Time Limit: 5 Sec   Memory Limit: 64 MB
Submit: 444   Solved: 219
[ Submit][ Status][ Discuss]

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

题意：

第一行输入两个数n和k表示有n个星球，你想要k号机器，而你一开始手上只有1号机器

接下来n行，每行两个数ai, bi表示第i个星球愿意拿bi号机器换ai号机器

问你最少需要换少次才能换到你想要的k号机器？

如果能换到输出这个答案+1，否则输出-1

每号机器当成一个点，每个ai和bi连一条ai到bi的有向边

求出1到k的最短路即可

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int bet[1005], len[52005];
int cnt, node[52005], nxt[52005], head[1005];
void Add(int x, int y, int z)
{
	node[++cnt] = y;
	nxt[cnt] = head[x];
	head[x] = cnt;
	len[cnt] = z;
}
void Dij()
{
	int i, v, now;
	memset(bet, 10, sizeof(bet));
	bet[1] = 0;
	priority_queue<pair<int, int> > q;
	q.push(make_pair(0, 1));
	while(1)
	{
		while(q.empty()==0 && -bet[q.top().second]<q.top().first)
			q.pop();
		if(q.empty())
			break;
		now = q.top().second;
		q.pop();
		for(i=head[now];i;i=nxt[i])
		{
			v = node[i];
			if(bet[v]>bet[now]+len[i])
			{
				bet[v] = bet[now]+len[i];
				q.push(make_pair(-bet[v], v));
			}
		}
	}
}
int main(void)
{
	int i, n, T, x, y;
	scanf("%d%d", &n, &T);
	memset(head, 0, sizeof(head));
	for(i=1;i<=n;i++)
	{
		scanf("%d%d", &x, &y);
		Add(x, y, 1);
	}
	Dij();
	if(bet[T]<=n)
		printf("%d\n", bet[T]+1);
	else
		printf("-1\n");
	return 0;
}