Catch That Cow(bfs)

Question

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.

Simple Input

5 17

Simple 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.

Code

#include 
#include 
#include 
using namespace std;
bool vis[100005];
int a, b;
struct node
{
    int x, step;
};
queueq;
bool OK(int t)
{
    if (t >= 0 && t <= 100000)
        return 1;
    else
        return 0;
}
int BFS()
{
    node start, mid, next;
    start.x = a;
    start.step = 0;
    vis[start.x] = true;
    if (start.x != b)
        q.push(start);
    while (!q.empty())
    {
        mid = q.front();
        q.pop();
        if (mid.x - 1 == b)
            return mid.step + 1;
        else if (OK(mid.x - 1) && !vis[mid.x-1])
        {
            next.x = mid.x - 1;
            next.step = mid.step + 1;
            q.push(next);
            vis[next.x] = true;//注意：入队后记得标记，否则内存会超；
        }
        if (mid.x + 1 == b)
            return mid.step + 1;
        else if (OK(mid.x + 1) && !vis[mid.x + 1])
        {
            next.x = mid.x + 1;
            next.step = mid.step + 1;
            q.push(next);
            vis[next.x] = true;
        }
        if (mid.x * 2 == b)
            return mid.step + 1;
        else if (OK(mid.x * 2) && !vis[mid.x * 2])
        {
            next.x = mid.x * 2;
            next.step = mid.step + 1;
            q.push(next);
            vis[next.x] = true;
        }
    }
    return 0;
}

int main()
{
    memset(vis, false, sizeof(vis));
    cin >> a >> b;
    cout << BFS() << endl;

    return 0;
}