【HDOJ-2717】Catch That Cow

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
 
思路:
当n>=k的时候只能后退一步 所以就是n-k
n<k的时候有三种 +1 -1 *2 BFS
 
参考代码
【HDOJ-2717】Catch That Cow
 1 #include <stdio.h>
 2 #include <string.h>
 3 int bfs(int n,int k);
 4 #define LEN 100001
 5 int step[LEN]={0};
 6 int mark[LEN]={0};
 7 int main()
 8 {
 9     int n,k;
10     while(scanf("%d%d",&n,&k)!=EOF)
11     {
12         if(n>=k)
13             printf("%d\n",n-k);
14         else
15             printf("%d\n",bfs(n,k));
16     }
17     return 0;
18 }
19 
20 int bfs(int n,int k)
21 {
22     memset(step,0,sizeof(step));
23     memset(mark,0,sizeof(mark));
24     int queue[LEN];
25     int i,front=0,rear=1;
26     queue[0]=n;
27     mark[queue[0]]=1;
28     while(front<rear)
29     {
30         for(i=0;i<3;i++)
31         {
32             if(queue[front]-1>=0&&!mark[queue[front]-1])
33             {
34                 mark[queue[front]-1]=1;
35                 step[queue[front]-1]=step[queue[front]]+1;
36                 queue[rear++]=queue[front]-1;
37                 if(queue[rear-1]==k)
38                     return  step[queue[front]-1];
39             }
40             if(queue[front]+1<=LEN&&!mark[queue[front]+1])
41             {
42                 mark[queue[front]+1]=1;
43                 step[queue[front]+1]=step[queue[front]]+1;
44                 queue[rear++]=queue[front]+1;
45                 if(queue[rear-1]==k)
46                     return step[queue[front]+1];
47             }
48             if(queue[front]*2<=LEN&&!mark[queue[front]*2])
49             {
50                 mark[queue[front]*2]=1;
51                 step[queue[front]*2]=step[queue[front]]+1;
52                 queue[rear++]=2*queue[front];
53                 if(queue[rear-1]==k)
54                     return step[queue[front]*2];
55             }
56         }
57         front++;
58     }
59 }
【HDOJ-2717】Catch That Cow

 

【HDOJ-2717】Catch That Cow

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