Catch That Cow

链接：https://ac.nowcoder.com/acm/contest/984/L
来源：牛客网
 

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?

输入描述:

Line 1: Two space-separated integers: N and K

输出描述:

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例1

输入

复制

5 17

输出

复制

4

说明

Farmer John starts at point 5 and the fugitive cow is at point 17.
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<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;
const int N=100010;
int n,k;
struct node{
	int x,step;
}; 
int map[N];
queue <node> Q;
int bfs(int x)
{
	while(!Q.empty())
	Q.pop();
	node a,next;
	a.x=x;
	a.step=0;
	map[x]=1;
	Q.push(a);
	while(!Q.empty())
	{
		a=Q.front();
		Q.pop();
		if(a.x==k)
			return a.step;
		for(int i=0;i<3;i++)
		{
			next=a;
			if(i==0)
				next.x++;
			else if(i==1)
				next.x--;
			else
				next.x*=2;	
			next.step++;
			if(next.x==k)
				return next.step;
			if(next.x>=0&&next.x<=N&&!map[next.x])
			{
				map[next.x]=1;	
				Q.push(next);	
			}
		
		} 
		
	}
}
int main()
{
	cin>>n>>k;
	cout<<bfs(n);
	return 0; 
}