暑假集训(1)第二弹 -----Catch the cow（Poj3278)

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.

 

 

问题分析：

简单bfs可以解答，分支为step+1；step-1；以及step*2. 第一次使用bfs以及容器；据说用容器比较慢，有时间想一想替代的方法0.0

 

 #include "iostream"
 #include "queue"

 using namespace std;
 char a[];
 void mset()
 {
     for (int i=;i<;i++)
          a[i] = '';
 }
 struct far
 {
     int w;
     int step;
 };
 int bfs(int n,int k)
 {
    if (n>=k)
         return n-k;
     queue<far> q;
     far fir,sec;
     fir.w = n;
     fir.step =;
     a[n] = '';
     q.push(fir);
     while (!(q.empty()))
     {
        sec=q.front();
        q.pop();
        for (int i=;i<=;i++)
        {
              switch (i)
               {
                 case  : fir.w=sec.w+; break;
                 case  : fir.w=sec.w-; break;
                 case  : fir.w=sec.w*; break;
               }
               fir.step=sec.step+;
               if (fir.w == k)   return fir.step;
               if (fir.w>= && fir.w<= && a[fir.w] == '' )
              {
                 a[fir.w] ='';
                 q.push(fir);
              }
        }
     }
 }
 int main()
 {
    int n,k;
    while (cin>>n>>k)
    {
        mset();
        cout<<bfs(n,k)<<endl;
    }
     return ;
 }