Catch That Cow

题目描述

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?

输入

Line 1: Two space-separated integers: N and K

输出

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

示例输入

5 17

示例输出

4

广度优先搜索练习

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<iostream>

 4 #include<queue>

 5 using namespace std;

 6 

 7 int vis[100100];

 8 struct node

 9 {

10     int date;

11     int step;

12 };

13 int bfs(int s, int t)

14 {

15     queue<node>que;

16     struct node tmp,ans;

17     int tx;

18     ans.date = s;

19     ans.step = 0;

20     que.push(ans);

21     vis[s] = 1;

22     while(!que.empty())

23     {

24         tmp = que.front();

25         que.pop();

26         tx = tmp.date-1;

27         if(tx < 0 || tx > 100000);

28         else if(!vis[tx])

29         {

30             vis[tx] = 1;

31             ans.date = tx;

32             ans.step = tmp.step+1;

33             que.push(ans);

34             if(ans.date == t)

35                 return ans.step;

36         }    

37         tx = tmp.date+1;

38         if(tx < 0 || tx > 100000);

39         else if(!vis[tx])

40         {

41             vis[tx] = 1;

42             ans.date = tx;

43             ans.step = tmp.step+1;

44             que.push(ans);

45             if(ans.date == t)

46                 return ans.step;

47         }

48         tx = tmp.date*2;

49         if(tx < 0 || tx > 100000);

50         else if(!vis[tx])

51         {

52             vis[tx] = 1;

53             ans.date = tx;

54             ans.step = tmp.step+1;

55             que.push(ans);

56             if(ans.date == t)

57                 return ans.step;

58         }

59     }

60 }

61 

62 int main()

63 {

64     int s,t;

65     scanf("%d %d",&s,&t);

66     if(s == t) 

67     {

68         printf("0\n");

69     }

70     else

71     {

72         memset(vis,0,sizeof(vis));

73         printf("%d\n",bfs(s,t));

74     }

75     return 0;

76 }

View Code