Catch That Cow

Catch That Cow

Time Limit:2000MS Memory Limit:65536K

Total Submit:118 Accepted:30

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.

问题描述：Farmer John要抓获逃亡的牛，他的开始位置N和牛的位置K是在同一条线上（0 ≤ N,K ≤ 100,000），Farmer John的追赶走有两种方式：步行和转送。

Walking：FJ每分钟移动的距离是X-1 or X+1 。

Teleporting： FJ每分钟移动的距离是2X 。

假设牛没有意识到追赶，原地不动，多久FJ能追赶上它？

输入： FJ和牛各自的位置

输出：抓到牛用的最短时间

Sample Input

5 17

Sample Output

4

Hint

FJ抓到牛用的最短时间路径: 5-10-9-18-17, 用时四分钟。

代码：

#include "stdafx.h"
#include<iostream>
#include<queue>
using namespace std;
#define M 100001
#define INF 10000000

// 定义一个结构体,pos表示位置，step表示所耗的分钟数。
struct POINT
{
int pos;
int step;
}now,m_next;

queue<POINT>Q;
bool visited[M];

int n,k;
// 计算追赶牛用的最短时间
int bfs()
{
// 栈清零
while(!Q.empty())
Q.pop();
// FJ所在的位置
now.pos=n;
now.step=0;
visited[now.pos]=true;
Q.push(now);
while(!Q.empty())
{
now=Q.front();
Q.pop();
m_next=now;
// 让程序更加有趣味一点
cout << "正在疯狂的抓牛！！" << endl;
for(int i=0;i<3;i++)
{
if(i==0)
{
cout << "前进一步！" << endl;
m_next.pos=now.pos+1;
}
if(i==1)
{
cout << "后退一步！" << endl;
m_next.pos=now.pos-1;
}
if(i==2)
{
cout << "前进两倍距离！" << endl;
m_next.pos=now.pos*2;
}
m_next.step=now.step+1;
if(m_next.pos==k)
return m_next.step;
if(m_next.pos<0||m_next.pos>M)
continue;
if(!visited[m_next.pos])
{
visited[m_next.pos]=true;
Q.push(m_next);
}
}
}
return INF;
}

int main()
{
while(cin>>n>>k)
{
memset(visited,false,sizeof(visited));
if(n<k)
cout<<bfs()<<endl;
if(n==k)
cout<<0<<endl;
if(n>k)
cout<<n-k<<endl;
}
}