Catch That Cow

算法：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

代码：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <queue>
using namespace std;
int a[400005];
struct dot
{
int x,step;
};
void bfs(int n,int m)
{   memset(a,0,sizeof(a));
queue<dot>que;
dot cur,loer;
cur.x=n;
cur.step=0;
a
=1;
que.push(cur);
while(que.size())
{
loer=que.front();
que.pop();
if(loer.x==m)
{
cout<<loer.step<<endl;
break;
}
for(int i=0;i<3;i++)
{
if(i==0)
cur.x=loer.x+1;
else if(i==1)
cur.x=loer.x-1;
else cur.x=loer.x*2;
if(cur.x>=0&&!a[cur.x]&&cur.x<200005)//一定要加上cur.x<200005
{
cur.step=loer.step+1;
a[cur.x]=1;
que.push(cur);
}
}
}
}
int main()
{
int n,m,i,j,k;
while(cin>>n>>m)
bfs(n,m);
return 0;
}