POJ3287(BFS水题)

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

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.

题意：

FJ要抓奶牛。

开始输入N（FJ的位置）K（奶牛的位置）。

FJ有三种移动方法：1、向前走一步，耗时一分钟。

2、向后走一步，耗时一分钟。

3、向前移动到当前位置的两倍N*2，耗时一分钟。

问FJ抓到奶牛的最少时间（奶牛不动）

解题思路：很明显我们求最短的时间就相当于求走的最少步数，所以我们就很容易想到用广搜BFS了，把每次FJ可以到的位置都压入队列，然后用一个vis数组标记FJ是否已经去过该位置了，再用一个step数组存储FJ到该位置所需要的最短时间即可。要注意的是，他的位置也有范围是0-100000，开始我就是因为右区间少了个=号，WA了好几次，所以还是要细心。难的做不出只好做简单的了。。。

附上代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string.h>
 4 #include<queue>
 5 using namespace std;
 6 int vis[100005];  //标记是否走过 
 7 int step[100005]; //储存到该处的最短时间 
 8 int n,k;
 9 queue<int> que;
10 int ans;
11 
12 int BFS()
13 {
14     que.push(n);
15     step[n]=0;
16     vis[n]=1;
17     while(que.size())
18     {
19         int p=que.front();
20         que.pop();
21         if(p==k) break;
22         for(int i=0;i<3;i++)
23         {
24             int dx;
25             if(i==0) dx=p-1;
26             if(i==1) dx=p+1;
27             if(i==2) dx=2*p;
28             if(dx>=0&&dx<=100000&&vis[dx]==0)
29             {
30                 que.push(dx);
31                 step[dx]=step[p]+1;
32                 vis[dx]=1;
33             }
34         
35         }
36     }
37     return step[k]; 
38 }
39 
40 int main()
41 {
42     while(cin>>n>>k)
43     {
44         memset(vis,0,sizeof(vis));
45         memset(step,0,sizeof(step));
46         ans=BFS();
47         cout<<ans<<endl;
48     }
49     return 0;
50  }

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