Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166    Accepted Submission(s): 3179

Problem 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.

 #include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
int b,e;
int Mintim;
struct node
{
int pos,t;
}k,tem;
int vis[+];
queue<node> s;
void bfs()
{
while(!s.empty())
s.pop();
k.pos=b,k.t=;
s.push(k);
while(!s.empty())
{
k=s.front();
s.pop();
if(k.pos==e)
{
Mintim=k.t;
return;
}
if(k.pos<||k.pos>||vis[k.pos]) continue;
vis[k.pos]=;
tem.t=k.t+;
tem.pos=k.pos+;
s.push(tem);
tem.pos=k.pos-;
s.push(tem);
tem.pos=k.pos*;
s.push(tem);
}
}
int main()
{
int i,j;
freopen("in.txt","r",stdin);
while(scanf("%d%d",&b,&e)!=EOF)
{
memset(vis,,sizeof(vis));
bfs();
printf("%d\n",Mintim);
}
return ;
}
 
 

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