C - Catch That Cow

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在一条数轴上，一位农夫从N点出发，使用步行或瞬移两种方式追赶位于K点的一头不动的逃逸奶牛。通过最优路径规划，本篇详细介绍了如何在最短时间内完成这一任务。

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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

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>
using namespace std;
#include<cstring>
const int N=1e5;
struct node{
	int num;
	int mm;
   node(int num,int mm){
   	this->num=num;
   	this->mm=mm;
   }
   node(){
   	
   }
};
bool flag[N*2];
struct node value[N*2];
int last=0;
void bfs(int start,int end){
   value[0]=node(start,0);
   flag[start]=true;
   ++last;
  for(int i=0;i<last;++i){
  	  int tnum=value[i].num;
  	  int tmm=value[i].mm;
  	    if(tnum==end) {
  	    	cout<<tmm;
  	    	break;
		  }
		  ++tmm;
		  if(!flag[tnum-1]){
		  
  	    value[last++]=node(tnum-1,tmm);
  	    flag[tnum-1]=true;
  	}
  	    if(tnum+1<=N&&!flag[tnum+1]){
		   	     value[last++]=node(tnum+1,tmm);
		   	     flag[tnum+1]=true;
		   	 }
				 	     if(tnum*2<=N&&!flag[tnum*2]){
						  
  	     value[last++]=node(tnum*2,tmm);
  	     flag[tnum*2]=true;
  	 }
  	
  	    
      
          
  }
	
}
int main(){
	int start,end;
	cin>>start>>end;
	memset(flag,false,sizeof(flag));
	bfs(start,end);
	
}