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

2014年4月12日1,3100

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.

题解

裸的bfs。。。最大值设为max(n,k)+1,就可以了,证明略。。