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POJ 3278 Catch That Cow C++版

POJ题解 Slyar 110浏览 0评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

最基础的广搜题,拿<queue>玩一下...

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...

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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6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7年前 (2010-04-18)回复