hd 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12304    Accepted Submission(s): 3822

Problem 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

  
  
  
  


  
  
  
  

   
   
#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
int s,e;
int vis[1000010];
struct node{
    int x;
    int step;
};
int bfs()
{
    node now,next;
    queue<node > q;
    now.x = s;
    now.step = 0;
    q.push(now);
    while(!q.empty())
    {
        next = q.front();
        q.pop();
        if(next.x == e)
                return next.step;
        now.x = next.x + 1;
        if(now.x >=0 && now.x <= 1000000 && !vis[now.x])
        {
            now.step = next.step + 1;
            vis[now.x] = 1;
            q.push(now);
        }
         now.x = next.x - 1;
        if(now.x >=0 && now.x <= 1000000 && !vis[now.x])
        {
            now.step = next.step + 1;
            vis[now.x] = 1;
            q.push(now);
        }
         now.x = next.x * 2;
        if(now.x >=0 && now.x <= 1000000&& !vis[now.x])
        {
            now.step = next.step + 1;
            vis[now.x] = 1;
            q.push(now);
        }
    }
    return -1;
}
int main()
{
    while(cin >> s >> e)
    {
        memset(vis,0,sizeof(vis));
        vis[s] = 1;
        int ans = bfs();
        cout << ans << endl;
    }
    return 0;
}