Poj 3278 Catch That Cow

1.Link:

http://poj.org/problem?id=3278

2.Content:

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 48245		Accepted: 15114

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

3.Method:

4.Code:

#include<iostream>
#include<queue>

using namespace std;

//(0 ≤ N ≤ 100,000)
#define MAX 200002
int a[MAX];
int main()
{
    int i,j;
    int x,y;
    int m,n;
    cin>>x>>y;
    queue<int> q;
    a[x]=1;
    q.push(x);
    while(!q.empty())
    {
        m=q.front();
        if(m==y) {cout<<a[m]-1;break;}
        else
        {
            n=m-1;
            if(n>=0&&n<MAX&&a[n]==0)
            {
                a[n]=a[m]+1;
                q.push(n);
            }
            n=m+1;
            if(n>=0&&n<MAX&&a[n]==0)
            {
               a[n]=a[m]+1;
               q.push(n);
            }
            n=m*2;
            if(n>=0&&n<MAX&&a[n]==0)
            {
               a[n]=a[m]+1;
               q.push(n);
            }
        }
        q.pop();
    }
    //system("pause");
    return 0;
}

5.Reference: