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.

 

 1 #include<stdio.h>

 2 #include<stdlib.h>

 3 #include<string.h>

 4 struct vode

 5 {

 6     int time,weizhi;

 7 };

 8 struct vode que[100001];

 9 int main()

10 {

11     int l,r,weizhi;

12     int n,k;

13     scanf("%d%d",&n,&k);

14     l=0,r=0;

15     weizhi=n;

16     que[r].weizhi=weizhi;

17     que[r].time=0;

18     r++;

19     int flag[100001]={0};

20     flag[weizhi]=1;

21     while(que[l].weizhi!=k)

22     {

23         weizhi=que[l].weizhi;

24         if(weizhi-1>=0&&weizhi-1<=100000&&flag[weizhi-1]==0)

25         {

26             que[r].weizhi=que[l].weizhi-1;

27             que[r].time=que[l].time+1;

28             flag[weizhi-1]=1;

29             r++;

30         }

31         if(weizhi+1>=0&&weizhi+1<=100000&&flag[weizhi+1]==0)

32         {

33             que[r].weizhi=que[l].weizhi+1;

34             que[r].time=que[l].time+1;

35             flag[weizhi+1]=1;

36             r++;

37         }

38         if(weizhi*2>=0&&weizhi*2<=100000&&flag[weizhi*2]==0)

39         {

40             que[r].weizhi=que[l].weizhi*2;

41             que[r].time=que[l].time+1;

42             flag[weizhi*2]=1;

43             r++;

44         }

45         l++;

46     }

47     printf("%d\n",que[l].time);

48     return 0;

49 }

View Code