poj 3278

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
简单bfs

#include<iostream>
#include<queue>
using namespace std;
struct Node
{
    int x,t;//当前的坐标 当前的时间
};
int v[1100111];
void bfs(int x,int y)
{
    Node now,next;
    now.x=x;now.t=0;
    queue<Node>q;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();q.pop();
        if(now.x==y)
            break;
        //共需3种要处理两种状态
        next.x=now.x*2;//1 倍增
        next.t=now.t+1;// 
        if(next.x>=0&&next.x<=1100100&&next.t<v[next.x])
        {
            v[next.x]=next.t;
            q.push(next);
        }

        next.x=now.x+1;
        next.t=now.t+1;//2 向右走
        if(next.x>=0&&next.x<=1100100&&next.t<v[next.x])
        {
            v[next.x]=next.t;
            q.push(next);
        }

        next.x=now.x-1;
        next.t=now.t+1;//3 向左走
    if(next.x>=0&&next.x<=1100100&&next.t<v[next.x])
        {
            v[next.x]=next.t;
            q.push(next);
        }
    
    }
    cout<<now.t<<endl;
}
int main()
{
    int x,y;

       while(cin>>x>>y)
       {
           if(x>=y)//如果x>y  那么只能向后走
               cout<<x-y<<endl;
           else
           {
       
       for(int i=1;i<=1100100;i++)//方便状态转移
           v[i]=999999999;
        bfs(x,y);
           }
       }
       
    
    return 0;
}