# POJ 3278 Catch That Cow C++版

OJ题解 6196浏览

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?

Input

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

Output

For each k given in the input, print one line containing ak to the output.

Sample Input

5 17

Sample Output

4

Slyar:USACO的题，给出2个数n和k，问从n经过+1或者-1或者*2能到达k的最小步数。分3个方向BFS，找到后树的深度就是最小步数了。注意n可以比k大，这时只有-1一种办法可以从n到达k，直接减就行了，还有要注意边界的判断...

### 网友最新评论 (3)

1. 哈哈哈，从你这学到好多东西啊，非常感谢
蜗牛13年前（2011-04-10）回复
2. 好久没看C++了，当初只学了一点皮毛。
网络人生14年前（2010-04-18）回复