zhber
有好多做过的题没写下来,如果我还能记得就补吧

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.

题意是有n个星球,在每个星球上你可以用a[i]物品换b[i]物品,要求用最少步数换到m。

初看连题目都没看懂……觉得好像很难的样子……看懂之后发现,这不是水题吗

转成有n条有向边,边权为1,求1到m的最短路

#include<cstdio>
#include<cstring>
struct edge{
	int to,next,v;
}e[100010];
int head[1010];
int dist[1010];
int ans[1010];
bool mrk[1010];
int q[500010];
int n,m,x,y,cnt,len,t,w=1;
inline void ins(int u,int v,int w)
{
	e[++cnt].to=v;
	e[cnt].v=w;
	e[cnt].next=head[u];
	head[u]=cnt;
}
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=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void spfa()
{
	memset(dist,127/3,sizeof(dist));
	q[1]=1;mrk[1]=1;dist[1]=0;
	while (t<w)
	{
		int now=q[++t];
		for (int i=head[now];i;i=e[i].next)
		  if (dist[e[i].to]>dist[now]+e[i].v)
		  {
		  	dist[e[i].to]=dist[now]+e[i].v;
		  	if (!mrk[e[i].to])
		  	{
		  		mrk[e[i].to]=1;
		  		q[++w]=e[i].to;
		  	}
		  }
		mrk[now]=0;
	}
}
int main()
{
	m=read();n=read();
	for (int i=1;i<=m;i++)
	  {
	  	x=read();y=read();
	  	ins(x,y,1);
	  }
	spfa();
	if (dist[n]>10000)dist[n]=-2;
	printf("%d\n",dist[n]+1);
	
}


posted on 2014-07-25 23:08  zhber  阅读(184)  评论(0编辑  收藏  举报