Catch That Cow+BFS

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4240    Accepted Submission(s): 1371

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


   

    

     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.
   
    
  

 

Source

USACO 2007 Open Silver

 

Recommend

teddy

 

思路：BFS

 

#include<iostream>
#include<queue>
using namespace std;
typedef struct infor
{
	int x,step;
}infor;
int n,k,Step[100002],min1;
queue<infor>q;
int vi[100002];
void BFS(){
	memset(vi,0,sizeof(vi));
	infor a;
	a.x=n;
	a.step=0;
	vi[n]=1;
	q.push(a);

	while(!q.empty()){
		infor b=q.front();
		q.pop();
		for(int i=0;i<3;i++){
			infor c=b;
			switch(i){
			case 0:{
						c.x+=1;
						c.step++;
						if(c.x<=100000&&!vi[c.x])
						{
							vi[c.x]=1;
							if(c.x==k)
								min1=min1>c.step?c.step:min1;
							q.push(c);
						}
						break;
				   }
			case 1:{
						c.x-=1;
						c.step++;
						if(c.x>=0&&!vi[c.x])
						{
							vi[c.x]=1;
							if(c.x==k)
								min1=min1>c.step?c.step:min1;
							q.push(c);
						}
						break;
				   }
			case 2:{
						c.x*=2;
						c.step++;
						if(c.x<=100000&&!vi[c.x])
						{
							vi[c.x]=1;
							if(c.x==k)
								min1=min1>c.step?c.step:min1;
							q.push(c);
						}
						break;
				   }
			}
		}
	}
	
}
int main()
{
	while(cin>>n>>k){
		min1=1000000;
		if(n==k)
			cout<<"0"<<endl;
		else
		{
			BFS();
			cout<<min1<<endl;
		}
	}
	return 0;
}