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 <iostream>#include <queue>#include <cstring>using namespace std;bool vis[100005];int a, b;struct node{    int x, step;};queue<node>q;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;}