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[100010];
void mset()
{
    for (int i=0;i<100000;i++)
         a[i] = '1';
}
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 =0;
    a[n] = '0';
    q.push(fir);
    while (!(q.empty()))
    {
       sec=q.front();
       q.pop();
       for (int i=1;i<=3;i++)
       {
             switch (i)
              {
                case 1 : fir.w=sec.w+1; break;
                case 2 : fir.w=sec.w-1; break;
                case 3 : fir.w=sec.w*2; break;
              }
              fir.step=sec.step+1;
              if (fir.w == k)   return fir.step;
              if (fir.w>=0 && fir.w<=100000 && a[fir.w] == '1' )
             {
                a[fir.w] ='0';
                q.push(fir);
             }
       }
    }
}
int main()
{
   int n,k;
   while (cin>>n>>k)
   {
       mset();
       cout<<bfs(n,k)<<endl;
   }
    return 0;
}