Catch That Cow（POJ-3278)

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.   AC:

#include
#include
#include
#include
#include

using namespace std;
const int maxn= 1e7+5;
int n,k;
int ans=0;
int vis[maxn];
typedef struct{
    int x;
    int step;
}node;
queueq;

void bfs(){
    node start;
    start.x=n;
    start.step=0;
    vis[n]=1;
    q.push(start);
    while(!q.empty()){
        node a;
        a=q.front();
        if(a.x==k){
            ans=a.step;
            break;
        }
        q.pop();
        if(a.x-1>=0&&!vis[a.x-1]){
           node b;
           b.x=a.x-1;
           b.step=a.step+1;
           vis[a.x-1]=1;
           q.push(b);
        }
        if(a.x>n>>k;
    memset(vis,0,sizeof(vis));

    bfs();
    // for(int i=0;i<30;i++){
    //     cout<