C : easy BFS (have a little trap)

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.

analysis:      pay attention to the " 2*x"     it maybe will make the array bounds.

code:

#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100005;
const int INF = 1000000;
int ans[maxn] = {0};
int main()
{
    int n,k;
    cin >> n >> k;
    for(int i=0;i<maxn;i++){
        ans[i] = INF;
    }
    queue<int> que;
    que.push(n);
    ans[n] = 0;
    while(que.size()){
        int temp = que.front();
        int md[3] = {1,-1,temp};
        que.pop();
        for(int i=0;i<3;i++){
            int d = temp + md[i];
            if(d < maxn){		// attention ! in order to prevent the array from crossing the border.
                if(ans[d] == INF){
                    que.push(d);
                    ans[d] = ans[temp] + 1;
                }
            }
        }
        if(ans[k] != INF){
            cout << ans[k] <<endl;
            break;
        }
    }
    return 0;
}