hdu2717(裸bfs广搜)

Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7304 Accepted Submission(s): 2308

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=2717
Problem 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: Two space-separated integers: N and K


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


Sample Input

5 17



Sample Output

4
Hint

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.

解题思路：

          一道很裸的bfs题。题意是给出farmer的点和cow的点，farmer可以选择每次+1、-1、*2这三种操作，问最少的操作次数可以使得farmer到达cow点。

          重新温习广搜。广搜对于求最优解或最短路很有帮助，虽然它的时间复杂度颇高，要把整棵树的所有路径都存下来，不过据说双向广搜（dbfs）可以把时间复杂度优化到O（n），加上哈希表处理的话，可以达到O（1）。

           本题考虑两种情况：1、 n < k时，有+1、*2两种走法；2、 n > k 时，只有-1。其实最后所求的最少次数相当于询问树节点为k的最短深度。

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