catch that cow (bfs 搜索的实际应用，和图的邻接表的bfs遍历基本上一样)

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 38263		Accepted: 11891

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.

 

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 struct vode
 {
     int time,weizhi;
 };
 struct vode que[];
 int main()
 {
     int l,r,weizhi;
     int n,k;
     scanf("%d%d",&n,&k);
     l=,r=;
     weizhi=n;
     que[r].weizhi=weizhi;
     que[r].time=;
     r++;
     int flag[]={};
     flag[weizhi]=;
     while(que[l].weizhi!=k)
     {
         weizhi=que[l].weizhi;
         if(weizhi->=&&weizhi-<=&&flag[weizhi-]==)
         {
             que[r].weizhi=que[l].weizhi-;
             que[r].time=que[l].time+;
             flag[weizhi-]=;
             r++;
         }
         if(weizhi+>=&&weizhi+<=&&flag[weizhi+]==)
         {
             que[r].weizhi=que[l].weizhi+;
             que[r].time=que[l].time+;
             flag[weizhi+]=;
             r++;
         }
         if(weizhi*>=&&weizhi*<=&&flag[weizhi*]==)
         {
             que[r].weizhi=que[l].weizhi*;
             que[r].time=que[l].time+;
             flag[weizhi*]=;
             r++;
         }
         l++;
     }
     printf("%d\n",que[l].time);
     return ;
 }