# 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?
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.

```#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
struct Point
{
int pos,step;
}start;
queue<Point> que;
int n=0,m=0;
bool vis;
int Bfs()
{
while(!que.empty())
{
Point temp=que.front();
que.pop();
if(temp.pos==m) return temp.step;
for(int i=1;i<=3;i++)
{
Point next_point=temp;
if(i==1) next_point.pos++;
else if(i==2) next_point.pos--;
else if(i==3) next_point.pos*=2;
next_point.step++;

if((next_point.pos>=0&&next_point.pos<=200000)&&!vis[next_point.pos])
{
que.push(next_point);
vis[next_point.pos]=1;
}
if(next_point.pos==m) return next_point.step;
}
}
}
int main()
{
cin>>n>>m;
start.pos = n;
start.step=0;
vis[start.pos]=1;
que.push(start);
printf("%d\n",Bfs());
return 0;
}
```