poj 3278 Catch That Cow BFS
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http://acm.pku.edu.cn/JudgeOnline/problem?id=3278
poj 3278 Catch That Cow
Time Limit:2000MS Memory Limit:65536K
Total Submit:2315 Accepted:530
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.
Source
USACO 2007 Open Silver
Description:
数轴上有两个数N和K,对N每次可以+1,-1,*2操作,问至少几次运算N能等于K
Solution:
简单基础题 考察基本的BFS RE了几次 数组不够大 放大到200001就可以AC了
1.循环队列
2.标记数组用于限定检查数的范围(N<=2*K&&N>0)
3.BFS层数记录每次记录这一层的最后一个结点在队列中的下标
4.if N>=K then output N-K
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