POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30924		Accepted: 9536

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

/* 功能Function Description:     POJ 3278 Catch That Cow
   开发环境Environment:          DEV C++ 4.9.9.1
   技术特点Technique:
   版本Version:
   作者Author:                    可笑痴狂
   日期Date:                      20120730
   备注Notes:                     深搜---队列
*/
#include<iostream>
#include<queue>
#define MAX 100001
using namespace std;

queue<int> q;
bool visit[MAX];
int step[MAX];       //记录步数的数组不能少

bool bound(int num)
{
    if(num<0||num>100000)
        return true;
    return false;
}

int BFS(int st,int end)
{
    queue<int> q;
    int t,temp;
    q.push(st);
    visit[st]=true;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        for(int i=0;i<3;++i) //三个方向搜索 
        {
            if(i==0)
                temp=t+1;
            else if(i==1)
                temp=t-1;
            else
                temp=t*2;
            if(bound(temp))         //越界
                continue;
            if(!visit[temp])
            {
                step[temp]=step[t]+1;
                if(temp==end)
                    return step[temp];
                visit[temp]=true;
                q.push(temp);
            }
        }
    }
}

int main()
{
    int st,end;
    while(scanf("%d%d",&st,&end)!=EOF)
    {
        memset(visit,false,sizeof(visit));
        if(st>=end)
            cout<<st-end<<endl;
        else
            cout<<BFS(st,end)<<endl;
    }
    return 0;
}