Catch That Cow

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal 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;}