抓牛问题

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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.

//借鉴网上的，代码很强大，也很简洁。思路就是利用广搜直接判断出到达某一位置最短步数

#include<cstdio>
#define N 100005

int u,v,x,f[N],q[N*2],h,t;
int main()
{
    scanf("%d%d",&u,&v);
    q[t++]=u;
    while(h<t)
    {
        u=q[h++];
        if(u==v)
        {
            printf("%d
",f[u]);
            break;
        }
        x=f[u]+1;
        if(u>0&&!f[u-1])
            f[q[t++]=u-1]=x;
        if(u<=N&&!f[u+1])
            f[q[t++]=u+1]=x;
        if(u*2<=N&&!f[u*2])
            f[q[t++]=u*2]=x;
    }
}