poj 3278 catch that cow BFS(基础水)
| Time Limit: 2000MS | ? | Memory Limit: 65536K |
| Total Submissions: 61826 | ? | Accepted: 19329 |
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 KOutput
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.Sample Input
5 17Sample Output
4Hint
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
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#include<stdio.h> #include<string.h> #include<queue> #include<iostream> #include<algorithm> using namespace std;轉(zhuǎn)載于:https://www.cnblogs.com/13224ACMer/p/4738717.html
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