Catch That Cow POJ

Catch That Cow

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 94224 Accepted: 29551

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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 <cstdio>#include <cstring>#include <iostream>#include <stack>#include <queue>using namespace std;int n,k;int book[120123],que[120123];int has(int n){    if(n<0||n>100000||book[n])        return 0;    else return 1;}int BFS(){    if(n == k) return 0;    int head=0,tail=1;    que[head] = n;    while(head<tail)    {        int x=que[head];        head++;        if(x-1 ==k || x+1 ==k || x*2 ==k)            return book[x]+1;        if(has(x-1))        {            book[x-1]=book[x]+1;            que[tail++]=x-1;        }        if(has(x+1))        {            book[x+1]=book[x]+1;            que[tail++]=x+1;        }        if(has(x*2))        {            book[x*2] = book[x]+1;            que[tail++]=2*x;        }    }    return -1;}int main(){    int i,j;    while(cin>>n>>k)    {        int ans = BFS();        cout<<ans<<endl;    }    return 0;}