B - Catch That Cow (抓牛)

B - Catch That Cow

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.

 

//用bfs，从当前位置开始，不断的试探，+1，-1，*2。直到找到牛，农民位置大于牛的时候，直接输出农民位置减牛的位置就行。

 

 #include <stdio.h>
 #include <string.h>
 #include <queue>
 using namespace std;  

 const int N = ;
 int map[N+];
 int n,k;

 struct node
 {
     int x,step;
 };  

 int check(int x)
 {
     if(x< || x>=N || map[x])
         return ;
     return ;
 }  

 int bfs(int x)
 {
     queue <node> Q;
     node a,next;

     a.x = x;
     a.step = ;
     map[x] = ;
     Q.push(a);

     while(!Q.empty())
     {
         a = Q.front();
         Q.pop();

         if(a.x == k)
             return a.step;
         next = a;
         //每次都将三种状况加入队列之中
         next.x = a.x+;
         if(check(next.x))
         {
             next.step = a.step+;
             map[next.x] = ;
             Q.push(next);
         }
         next.x = a.x-;
         if(check(next.x))
         {
             next.step = a.step+;
             map[next.x] = ;
             Q.push(next);
         }
         next.x = a.x*;
         if(check(next.x))
         {
             next.step = a.step+;
             map[next.x] = ;
             Q.push(next);
         }
     }
     return ;
 }

 int main()
 {
     int ans;
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         memset(map,,sizeof(map));
         if (n>k) ans=n-k;
         else ans = bfs(n);
         printf("%d\n",ans);
     }
     return ;
 }