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

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.

讲解:一农夫追牛,牛很笨,他就在原地等着农夫来抓他,并且这是一条直线,农夫很容易就能找到它的,然而农夫却只有三种选择,退一步,走一步,或者走到当前位置的2倍;于是乎我们可以用搜索来解决;

代码如下:

 #include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int map[];
struct T
{
int x,step;
};
int dfs(T now)
{queue< T >q;
T end;
q.push(now);
while(!q.empty())
{
end=q.front();
q.pop();
map[end.x]=;
if(end.x==m)
{return end.step;}
//如果不是走一步一判断很容易超出内存的
now.x=end.x+;//前进一步,存入队列;
if(end.x>= && end.x<= && map[now.x]==)
{
now.step=end.step+;
map[now.x]==;
q.push(now);
}
now.x=end.x-;//后退一步
if(end.x>= && end.x<= && map[now.x]==)
{
now.step=end.step+;
map[now.x]==;
q.push(now);
}
now.x=end.x*;//前进2倍的位置
if(end.x>= && end.x<= && map[now.x]==)
{
now.step=end.step+;
map[now.x]==;
q.push(now);
}
}
return ;
}
int main()
{
T now;
while(cin>>n>>m)
{
if(n>=m)//如果n大于m则只能后退了;
{
cout<<n-m<<endl;continue;
}
else
{
memset(map,,sizeof(map));
now.x=n;now.step=;
int mm=dfs(now);
cout<<mm<<endl;
}
}
return ;
}

hdoj 2717 Catch That Cow的更多相关文章

  1. hdoj 2717 Catch That Cow【bfs】

    Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  2. Hdoj 2717.Catch That Cow 题解

    Problem Description Farmer John has been informed of the location of a fugitive cow and wants to cat ...

  3. HDU 2717 Catch That Cow (bfs)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2717 Catch That Cow Time Limit: 5000/2000 MS (Java/Ot ...

  4. HDU 2717 Catch That Cow --- BFS

    HDU 2717 题目大意:在x坐标上,农夫在n,牛在k.农夫每次可以移动到n-1, n+1, n*2的点.求最少到达k的步数. 思路:从起点开始,分别按x-1,x+1,2*x三个方向进行BFS,最先 ...

  5. HDU 2717 Catch That Cow(常规bfs)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2717 Catch That Cow Time Limit: 5000/2000 MS (Java/Oth ...

  6. HDU 2717 Catch That Cow(BFS)

    Catch That Cow Farmer John has been informed of the location of a fugitive cow and wants to catch he ...

  7. HUD 2717 Catch That Cow

    Catch That Cow Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...

  8. hdu 2717:Catch That Cow(bfs广搜,经典题,一维数组搜索)

    Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  9. hdu 2717 Catch That Cow(广搜bfs)

    题目链接:http://i.cnblogs.com/EditPosts.aspx?opt=1 Catch That Cow Time Limit: 5000/2000 MS (Java/Others) ...

随机推荐

  1. Oracle 去重查询

      Oracle 去重查询 CreateTime--2018年2月28日15:38:45 Author:Marydon (一)使用distinct --查询指定区间内表停诊字段的值 SELECT DI ...

  2. eclipse svn插件

      eclipse svn插件 CreateTime--2018年4月22日23:09:42 Author:Marydon 下载地址:eclipse svn插件 将svn插件目录直接拷贝到eclips ...

  3. JQuery 在线参考手册

    官方JQuery API  http://api.jquery.com/ 在线参考手册    http://www.w3school.com.cn/jquery/index.asp 在线参考手册1   ...

  4. Oracle执行SQL报错ORA-00922

    问题描述: 对Oracle数据库执行序列化脚本出错,ora-00922 missing or invalid option  #无效的选项 问题解决: 对于set define off --这个是sq ...

  5. 原生php如何获取当前页面的url

    原生php如何获取当前页面的url? //php获取当前访问的完整url地址 function get_current_url(){ $current_url='http://'; if(isset( ...

  6. 如何解决普通用户使用sudo找不到命令

    一.在linux的普通用户下,要使用root权限的命令需要使用sudo [dev@dev1 client_api]# sudo git pull origin develop sudo: git: c ...

  7. POJ--Strange Way to Express Integers

    Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 8370   ...

  8. 推荐系统之矩阵分解及其Python代码实现

    有如下R(5,4)的打分矩阵:(“-”表示用户没有打分) 其中打分矩阵R(n,m)是n行和m列,n表示user个数,m行表示item个数 那么,如何根据目前的矩阵R(5,4)如何对未打分的商品进行评分 ...

  9. App上架重磅通知:App Store安全新规17年1月生效

    作者:沙铭 来源:公众号 沙铭世界观 ID:mobview 做推广的也许并不了解什么是ATS(App Transport Security),不过这却是一个定时炸弹,引爆点在2016年底,后果就是你不 ...

  10. ios NSURLSession使用说明及后台工作流程分析

    NSURLSession是iOS7中新的网络接口,它与咱们熟悉的NSURLConnection是并列的.在程序在前台时,NSURLSession与NSURLConnection可以互为替代工作.注意, ...