POJ-3278-Catch That Cow(bfs)
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?
Line 1: Two space-separated integers: N and K
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
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.
农场主的牛不见了，主人和牛在一条直线上，且牛没有新的目标，它不会走动，主人的位置是你 n，牛的位置是 k，主人可以有三种走路的方法，右左（距离+-1），闪现（距离+x,x 为当前位置），每走一步，一分钟，问几分钟主人能找到牛。bfs 搜索方向即为三个“方向”。搜索所有走法；