ayit第十四周训练题d题

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.

 

#include
#include
int vis[1001010];
int n, k;
struct node
{
    int x;
    int step;
}q[1001010];
int bfs()
{
    int i, head, tail;
     struct node now, next;
    head=0;
    tail=0;
    q[tail].x = n;
    q[tail].step = 0;
    tail++;
    vis[n] = 1;
    while(head < tail)
    {
        now = q[head];
        head++;
        for(i=0; i<3; i++)
        {
            if(i==0)
                next.x=now.x-1;
            else if(i == 1)
                next.x = now.x +1;
            else
                next.x=2*now.x;
            if(next.x<0|| next.x >=1001010)
                continue;
            if(!vis[next.x])
            {
                vis[next.x] = 1;
                next.step = now.step + 1;
                q[tail].x = next.x;
                q[tail].step = next.step;
                tail ++;
                if(next.x == k)
                    return next.step;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d", &n,&k))
    {
        memset(vis, 0, sizeof(vis));
        if(n >= k)
            printf("%d\n", n-k);
        else
            printf("%d\n", bfs());
    }
    return 0;
}

题意  已知牛和人的位置，人单位时间只能左或右走一步，或者走现在的两倍距离（在数轴上，且牛不动），问什么时候能追上             牛

思路   广搜题 一直往深处走，直到找到解或者走不下去为止