Catch That Cow

 

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 77940		Accepted: 24626

 

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

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.

Source

USACO 2007 Open Silver

#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 200005;
int que[maxn],hash[maxn];
int judge(int i)
{
    if(i > maxn || i < 0 || hash[i])
        return 0;
    return 1;
}
int bfs(int n,int k)
{
    if(n == k)
        return 0;
    int x,front = 0,base = 0;
    que[front++] = n;
    while(base < front)
    {
        x = que[base++];
        if(x-1 == k || x+1 == k || 2*x ==k)
        {
            return hash[x]+1;
        }
        if(judge(x-1))
        {
            hash[x-1] = hash[x]+1;
            que[front++] = x-1;
        }
        if(judge(x+1))
        {
            hash[x+1] = hash[x]+1;
            que[front++] = x+1;
        }
        if(judge(2*x))
        {
            hash[2*x] = hash[x]+1;
            que[front++] = 2*x;
        }
    }
    return -1;
}
int main()
{
    int n,k;
    while(cin >> n >> k)
    {
        memset(hash,0,sizeof(hash));
        int ans;
        ans = bfs(n,k);
        cout << ans << endl;
    }
    return 0;
}