poj3278-Catch That Cow

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.

Example Input

5 17

Example 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>
#include <cstring>
using namespace std;
int book[1000005];
int n,k;
bool judge(int x)
{
if(x<0||x>=1000000||book[x])
return 0;
return 1;
}
struct node
{
int step;
int point;
}t,v;
void bfs(int n)
{
queue<node>q;
t.point=n;
t.step=0;
q.push(t);
book
=1;
while(!q.empty())
{
v=q.front();
q.pop();
if(v.point==k)
{
cout<<v.step<<endl;
return ;
}
t=v;
t.point=v.point+1;
if(judge(t.point))
{
t.step=v.step+1;
book[t.point]=1;
q.push(t);
}
t.point=v.point-1;
if(judge(t.point))
{
t.step=v.step+1;
book[t.point]=1;
q.push(t);
}
t.point=v.point*2;
if(judge(t.point))
{
4000
t.step=v.step+1;
book[t.point]=1;
q.push(t);
}
}

}
int main(int argc, char const *argv[])
{
scanf("%d%d",&n,&k);
bfs(n);
return 0;
}