題目連結：傳送門

Problem 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

題目大意：給出兩個座標s，w。s在數軸上的行走規律是可以向左走一步，向右走一步，或跳到是當前座標位置的2倍的地方，而且這三種行走情況所消耗的時間都是一秒。求s到達w的最短時間。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int m[1000010];
int s,w;
struct nm
{
    int nat,tim;
};
int bfs()
{
    int a,b,c,d;
    nm next,tur;
    queue<nm>n;
    tur.nat=s;
    tur.tim=0;
    n.push(tur);
    m[s]=1;
    while(!n.empty())
    {
        tur=n.front();
        n.pop();
        if(tur.nat==w)
            return tur.tim;
        if(m[tur.nat-1]==0&&(tur.nat-1)>=0&&(tur.nat-1)<=100010)
        {
            next.nat=tur.nat-1;
            next.tim=tur.tim+1;
            n.push(next);
            m[tur.nat-1]=1;
        }
        if(m[tur.nat+1]==0&&(tur.nat+1)>=0&&(tur.nat+1)<=100010)
        {
            next.nat=tur.nat+1;
            next.tim=tur.tim+1;
            n.push(next);
            m[tur.nat+1]=1;
        }
        if(m[tur.nat*2]==0&&(tur.nat*2)>=0&&(tur.nat*2)<=100010)
        {
            next.nat=tur.nat*2;
            next.tim=tur.tim+1;
            n.push(next);
            m[tur.nat*2]=1;
        }
    }
    return -1;
}
int main()
{
    int a;
    while(cin>>s>>w)
    {
        memset(m,0,sizeof(m));
        a=bfs();
        cout<<a<<endl;
    }
    return 0;
}