Catch That Cow POJ - 3278

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

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．

    他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有

两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．

    那么，约翰需要多少时间抓住那只牛呢？

代码

1. #include <iostream>  
2. #include <cstdio>  
3. #include <cstring>  
4. #include <algorithm>  
5. #include <queue>  
6. #include <vector>  
7. using namespace std;  
8. const int MAXN=500001;  
9. bool visit[MAXN];  
10. vector <int> Q;  
11. struct xx  
12. {  
13.     int s,cnt;  
14. }a[MAXN];  
15. int main(int argc, char *argv[])  
16. {  
17.     int n,k,head,tail,d=0,zt,i;  
18.     scanf("%d%d",&n,&k);  
19.     a[1].s=n;  
20.     head=1,tail=1;  
21.     while(head<=tail)  
22.     {  
23.         zt=tail;  
24.         for(i=head;i<=tail;i++)  
25.         {  
26.             if(a[i].s==k) {  
27.                 printf("%d\n",a[i].cnt);  
28.                 return 0;  
29.             }  
30.             if(a[i].s+1<MAXN)  
31.             if(visit[a[i].s+1]==false)   
32.             {  
33.                 zt++;  
34.                 a[zt].s=a[i].s+1;  
35.                 a[zt].cnt=a[i].cnt+1;  
36.                 visit[a[zt].s]=true;  
37.             }  
38.             if(a[i].s-1>=0)  
39.             if(visit[a[i].s-1]==false)   
40.             {  
41.                 zt++;  
42.                 a[zt].s=a[i].s-1;  
43.                 a[zt].cnt=a[i].cnt+1;  
44.                 visit[a[zt].s]=true;  
45.             }  
46.             if(a[i].s*2<MAXN)  
47.             if(visit[a[i].s*2]==false)   
48.             {  
49.                 zt++;  
50.                 a[zt].s=a[i].s*2;  
51.                 a[zt].cnt=a[i].cnt+1;  
52.                 visit[a[zt].s]=true;  
53.             }  
54.             head=tail+1;  
55.             tail=zt;  
56.         }  
57.     }  
58.     return 0;  
59. }  

扫描二维码关注公众号，回复： 183468 查看本文章