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

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.

题解：

        抓住那头牛，农夫和牛在一个数轴上，但是有三种移动方式，就可以把这三种移动方式，当作方向。

        就相当于是一维的bfs，拯救天使是二维的，下个题是三维的。

        emmmm

        画个图，理解一下bfs.，小菜刚开始在csdn上搜题解，一直纳闷为啥不用比较就是最小的😓😓

        如果不是真菜，谁愿意混吃混喝等死呢

        如图：绿色表示搜索到每一步的步数，

                    圈圈里是搜索到的数字，也就是农夫的当前位置，

                    然后灰色的圈表示它已经被标记过，不再继续搜索，

                    同一颜色的圈圈的数字的步数是一样的

                    也就是说，将位置5 用三个方式搜索一下，从位置5走到位置4，6，10的步数为1，然后再走位置10相邻的三个位置，就是1+1=2.

                    啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊，菜是原罪。

        

代码：

#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <string.h>
#include <iostream>
#include <vector>
using namespace std;
const int maxn=100001;
bool vis[maxn]; int step[maxn]; queue <int> q;
int bfs(int n,int k)
{
        int first,next;
        q.push(n);
        step[n]=0;
        vis[n]=1;
        while(!q.empty())
        {
                first=q.front();
                q.pop();
                for(int i=0;i<3;i++)
                {
                        if(i==0) next=first-1;
                        else if(i==1) next=first+1;
                        else next=first*2;
                        if(next<0||next>=maxn) continue;
                        if(vis[next]==0)
                        {
                                q.push(next);
                                step[next]=step[first]+1;
                                vis[next]=1;
                        }
                        if(next==k) return step[next];
                }
        }
}
int main()
{
        int n,k;
        while(cin>>n>>k)
        {
                memset(step,0,sizeof(step));
                memset(vis,0,sizeof(vis));
                while(!q.empty()) q.pop();
                if(n>=k) cout<<n-k<<endl;
                else cout<<bfs(n,k)<<endl;
        }
return 0;

}