[ACM] hdu 2717 Catch That Cow (BFS）

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
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.
 

 

Source

USACO 2007 Open Silver

解题思路：

从位置n到位置k根据规则最少走几步可以到达，规则是从n可以走到n-1,可以走到n+1,也可以走到n*2.用bfs广搜来做，vis[]用来记录是否访问，hash[]用来记录走的步数。

遇到的问题有数组越界，一定要先判断n*2,n-1,n+1是否越界，还有问题就是步数的保存，因为很多节点都是同一个步数下的状态，因为每一步根据规则可以衍生出三种状态。所以用hash[ q.front() 下一个状态]=hash[q.front()]+1.用这种方法来记录步数，最后直接输出hash[k]就可以了。

代码：

[cpp] view plaincopy

1. #include <iostream>  
2. #include <string.h>  
3. #include <queue>  
4. using namespace std;  
5. const int len=100000;  
6. int hash[len+10];//hash[i]用来保存走到位置i时是第几步  
7. bool vis[len+10];  
8. int n,k;  
9.   
10. void bfs()  
11. {  
12.     memset(vis,0,sizeof(vis));  
13.     queue<int>q;  
14.     q.push(n);  
15.     vis[n]=1;  
16.     hash[n]=0;  
17.     while(!q.empty())  
18.     {  
19.         int temp=q.front();  
20.         q.pop();  
21.         if(temp+1==k||temp-1==k||temp*2==k)//找到  
22.         {  
23.             hash[k]=hash[temp]+1;  
24.             break;  
25.         }  
26.         if(temp-1>=0&&!vis[temp-1])  
27.         {  
28.             q.push(temp-1);  
29.             vis[temp-1]=1;  
30.             hash[temp-1]=hash[temp]+1;  
31.         }  
32.         if(temp+1<=len&&!vis[temp+1])  
33.         {  
34.             q.push(temp+1);  
35.             vis[temp+1]=1;  
36.             hash[temp+1]=hash[temp]+1;  
37.         }  
38.         if(temp*2<=len&&!vis[temp*2])  
39.         {  
40.             q.push(temp*2);  
41.             vis[temp*2]=1;  
42.             hash[temp*2]=hash[temp]+1;  
43.         }  
44.     }  
45. }  
46.   
47. int main()  
48. {  
49.     while(cin>>n>>k)  
50.     {  
51.         if(n==k)  
52.         {  
53.             cout<<0<<endl;  
54.             continue;  
55.         }  
56.         else  
57.             bfs();  
58.         cout<<hash[k]<<endl;  
59.     }  
60.     return 0;  
61. }