2014.6.14模拟赛【bzoj1646】[Usaco2007 Open]Catch That Cow 抓住那只牛

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 <= N <= 100,000) on a number line and the cow is at a point K (0 <= K <= 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting. * Walking: FJ can move from any point X to the points X-1 or X+1 in a single minute * Teleporting: FJ can move from any point X to the point 2*X in a single minute. If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

    农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

    他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

    那么，约翰需要多少时间抓住那只牛呢？

Input

* Line 1: Two space-separated integers: N and K

    仅有两个整数N和K.

Output

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

    最短的时间．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4

OUTPUT DETAILS:

The fastest way for Farmer John to reach the fugitive cow is to

move along the following path: 5-10-9-18-17, which takes 4 minutes.

好裸的bfs……n的规模才10w

原来还想的是二进制dp，结果发现我在没事找事……

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m,head,tail=1;
int dist[500001];
int q[500001];
inline bool mark(int x)
{
	return !(x<0||x>max(2*m+1,n+1));
}
int main()
{
	freopen("catchcow.in","r",stdin);
	freopen("catchcow.out","w",stdout);
	scanf("%d%d",&n,&m);
	memset(dist,-1,sizeof(dist));
	q[1]=n;dist[n]=0;
	while (head<tail)
	{
		head++;
		int now=q[head]-1;
		if(mark(now)&&dist[now]==-1)
		{
			q[++tail]=now;
			dist[now]=dist[q[head]]+1;
		}
		now=q[head]+1;
		if(mark(now)&&dist[now]==-1)
		{
			q[++tail]=now;
			dist[now]=dist[q[head]]+1;
		}
		now=q[head]*2;
		if(mark(now)&&dist[now]==-1)
		{
			q[++tail]=now;
			dist[now]=dist[q[head]]+1;
		}
		if(dist[m]!=-1) {printf("%d",dist[m]);return 0;}
	}
}