Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

 

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.

发现广度优先搜索适合什么题呢？就是那种在每个点都给你几个选择，然后问你最短路径的问题，对，就是这样，这样的题目最适合广度优先搜索。

这次的这个就是，每次Farmer有三个选择，然后求最短到达目的地的行程。简直是广搜的模板题。

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #pragma warning(disable:4996)using namespace std;int color[1000005];int dis[1000005];queue
           
             q;int main(){ //freopen("i.txt","r",stdin); //freopen("o.txt","w",stdout); int N,K; cin>>N>>K; if(N==K) { cout<<0<
            
             =0) { color[N-1]=1; dis[N-1]=dis[N]+1; q.push(N-1); } if(color[N+1]==0) { color[N+1]=1; dis[N+1]=dis[N]+1; q.push(N+1); } if(color[2*N]==0 && 2*N<=100000) { color[2*N]=1; dis[2*N]=dis[N]+1; q.push(2*N); } } } return 0;}
            
           
          
         
        
       
      
     
    

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