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.

Sample Input

5 17

Sample Output

题意描述：

把一个数变成另一个数，可以通过加1或减1或者变成现在数的2倍，求最少的步数

程序代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
struct data{int x;int y;int step;
};
int book[200010];
int main()
{int a,b,k;struct data A,B,next;queue<data>que;while(scanf("%d%d",&a,&b)!=EOF){while (!que.empty()){que.pop();}A.x=a;A.step=0;memset(book,0,sizeof(book));book[A.x]=1;que.push(A);while(!que.empty()){B=que.front();if(B.x==b){printf("%d\n",B.step);break;}que.pop();for(k=0;k<3;k++){next=B;if(k==0)next.x=next.x+1;if(k==1)next.x=next.x-1;if(k==2)next.x=next.x*2;next.step++;if(next.x>=0&&next.x<=200000&&book[next.x]==0){book[next.x]=1;que.push(next); }}}}return 0;
}

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