HDU2717-Catch That Cow （BFS入门）

题目传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2717

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14880    Accepted Submission(s): 4495

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.

 

题意：

从a走到b，你可以向前走一步，向后走一步或者走a*2步，求最少多少步

 

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 using namespace std;
 int v[],t[];
 int main()
 {
     int n,a,b,i,f;
     while(scanf("%d%d",&a,&b)!=EOF)
     {
         memset(t,,sizeof(t));
         memset(v,,sizeof(v));
         queue<int > q;
         q.push(a);
         v[a]=;
         if(a==b)
         {
             printf("0\n");
             continue;
         }
         while(!q.empty())
         {
             int ll=q.front(),r=ll-,l=ll+,z=ll*;
             q.pop();
             //printf("ll=%d\n",ll);
             if(z>=&&z<=&&!v[z])
             {
                 q.push(z);
                 v[z]=v[ll]+;
             //    printf("v[z]=%d z=%d\n",v[z],z);
             }
             if(l>=&&l<=&&!v[l])
             {
                 q.push(l);
                 v[l]=v[ll]+;
                 //    printf("v[l]=%d l=%d\n",v[l],l);
             }
             if(r>=&&r<=&&!v[r])
             {
                 q.push(r);
                 v[r]=v[ll]+;
                 //    printf("v[r]=%d r=%d\n",v[r],r);
              }

              if(l==b||r==b||z==b)
              {

                  break;
              }

         }

         printf("%d\n",v[b]-);

     }
     return ;

  }