Catch That Cow

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.

题意就是你通过三种操作抓到牛，没办法只能一次一次的压入队列，然后把有用的点留下来，没用的就放弃。

看代码可能更清楚。

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int N = 1000000;
bool book[N+10];
int n,m;
struct node
{
    int x,y;
};
int cm(int x)
{
    if(x<0||x>=N||book[x])
        return 0;
    return 1;
}
int cn()
{
    int i;
    queue<node> Q;
    node st,en;
    st.x = n;
    st.y= 0;
    book[st.x]=1;
    Q.push(st);
    while(!Q.empty())
    {
        st = Q.front();
        Q.pop();
        if(st.x == m)
            return st.y;
        en= st;
        en.x = st.x+1;
        if(cm(en.x))
        {
            en.y= st.y+1;
            book[en.x] = 1;
            Q.push(en);
        }
        en.x = st.x-1;
        if(cm(en.x))
        {
            en.y= st.y+1;
            book[en.x] = 1;
            Q.push(en);
        }
        en.x = st.x*2;
        if(cm(en.x))
        {
            en.y= st.y+1;
            book[en.x] = 1;
            Q.push(en);
        }
    }
    return -1;
}
int main()
{
    int sum;
    while(~scanf("%d%d",&n,&m))
    {
        memset(book,0,sizeof(book));
        sum = cn();
        printf("%d\n",sum);
    }
    return 0;
}