Catch That Cow 

HDU - 2717 

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.

输入

第 1 行：两个空格分隔的整数：N 和 K

输出

第 1 行：Farmer John 抓住逃亡母牛所需的最少时间（以分钟为单位）。

Sample Input

5 17

Sample Output

4

这道题 和A strange lift（奇怪的电梯）  HDU - 1548 差不多，道理都一样，都是利用队列和bfs                                       ????

有兴趣的话可以做一下⬆题，下面是????题的我的博客

https://blog.csdn.net/m0_58245389/article/details/119240688

本题AC

#include<stdio.h>
#include<queue>
#include<math.h>
#include<string.h>
#include<iostream>
int v[100010];
using namespace std;
struct Node
{
int space,step;
};
int main(void)
{
int n,m,b;
Node now,next;
queue<Node>Q;
while(cin>>n>>m)
{
memset(v,0,sizeof(v));
while(!Q.empty())
{
Q.pop();
}
now.space=n;
now.step=0;
Q.push(now);
v[now.space]=1;
while(!Q.empty())//分三路走
{
now=Q.front();
Q.pop();
if(now.space==m)
{
printf("%dn",now.step);
break;
}
b=now.space*2;
if(b<=100000&&!v[b])
{
v[b]=1;
next.space=b;
next.step=now.step+1;
Q.push(next);
}
b=now.space+1;
if(b<=100000&&!v[b])
{
v[b]=1;
next.space=b;
next.step=now.step+1;
Q.push(next);
}
b=now.space-1;
if(b>=0&&!v[b])
{
v[b]=1;
next.space=b;
next.step=now.step+1;
Q.push(next);
}
}
//printf("%dn",);
}
return 0;
}

最后

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