Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 61317		Accepted: 19155

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

 

 

RE: 坐标轴上有两个点（a、b）, 求 a 接近 b 所需走的最小步数；

　　有三种走法，（x-1)、 （x+1）、（2*x）;

　　a > b  → → 输出；

　　else :: Bfs;

　　

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 using namespace std; 6  7 const int maxn = 100000 + 5; 8 int vis[maxn], cnt[maxn]; 9 queue 
          
            q;10 11 void Bfs()12 {13     int nx;14     while(!q.empty())15     {16         int x = q.front();  q.pop();17         for(int i = 0; i < 3; i++)    //三种走法； 18         {19             if(i == 0) nx = x + 1;20             else if(i == 1) nx = x - 1;21             else nx = 2 * x;22             if(nx >= 0 && nx <= 100000 && !vis[nx])  //遍历到最后找最优？::剪枝？ 23             {24                 vis[nx] = 1;25                 cnt[nx] = cnt[x] + 1;26                 q.push(nx);     27             } 28         }     29     }30 }31 32 int main()33 {34     int m, n;35     while(~scanf("%d %d", &m, &n))36     {37         if(m > n)38         {39             printf("%d\n", m - n);40             return 0;41         }42         memset(vis, 0, sizeof(vis));43         vis[m] = 1;44         cnt[m] = 0;45         q.push(m);46         Bfs();47         printf("%d\n", cnt[n]);48     } 49     return 0;50 }