# 描述

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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来,他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有两种办法移动，步行和瞬移：步行每秒种可以让约翰从x处走到x+1或x-1处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．那么，约翰需要多少时间抓住那只牛呢？

# 输入输出格式

## 输入

* Line 1: Two space-separated integers: N and K
仅有两个整数N和K.

## 输出

* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
最短的时间．

# 输入输出样例

## 输入样例1

5 17

## 输出样例1

4


# 解题思路

这里特判n>=k的情况，然后搜索三种方式，再剪个枝就行了。

# 题解

 1 #include<bits/stdc++.h>
2 using namespace std;
3 int n,k;
4 queue<int> q;
5 int flag[1000001];//标记
6 void bfs()
7 {
8     q.push(n);
9     flag[n]=1;
10     while(!q.empty())
11     {
13         q.pop();
16         {
19         }
21         {
24         }
26         {
29         }
30     }
31     cout<<flag[k]-1;
32 }
33 int main()
34 {
35     cin>>n>>k;
36     if(n>=k)//特判，只能一步一步向回走
37     {
38         cout<<n-k;
39         return 0;
40     }
41     bfs();
42     return 0;
43 }
44     

posted @ 2019-07-17 12:23  华恋~韵  阅读(83)  评论(0编辑  收藏  举报