问题描述:

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.

解题思路:

bfs,三个方向搜索,x=x+1,x=x-1,x=x*2,最先搜索到的就是用时最短的。

代码:

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
using namespace std;
int n,k;
int a[]; struct node
{
int x;
int step;
}now,net; int bfs(int x)
{
queue<node> q;
now.x=x;
now.step=;
q.push(now);
while(q.size())
{
now=q.front();
q.pop();
//三种情况
if(now.x==k)return now.step;
net.x=now.x+;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x-;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x*;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
}
return -;
} int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(a,,sizeof(a));
a[n]=;
int ans=bfs(n);
printf("%d\n",ans);
}
return ;
}

Catch That Cow (BFS广搜)的更多相关文章

  1. Catch That Cow(广搜)

    个人心得:其实有关搜素或者地图啥的都可以用广搜,但要注意标志物不然会变得很复杂,想这题,忘记了标志,结果内存超时: 将每个动作扔入队列,但要注意如何更简便,更节省时间,空间 Farmer John h ...

  2. poj 3278 Catch That Cow (广搜,简单)

    题目 以前做过,所以现在觉得很简单,需要剪枝,注意广搜的特性: 另外题目中,当人在牛的前方时,人只能后退. #define _CRT_SECURE_NO_WARNINGS //这是非一般的最短路,所以 ...

  3. HDU2717 Catch That Cow 【广搜】

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

  4. Catch That Cow 经典广搜

    链接:http://poj.org/problem?id=3278 题目: Farmer John has been informed of the location of a fugitive co ...

  5. hdu 1242:Rescue(BFS广搜 + 优先队列)

    Rescue Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submis ...

  6. hdu 1195:Open the Lock(暴力BFS广搜)

    Open the Lock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tot ...

  7. 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 ...

  8. BFS广搜题目(转载)

    BFS广搜题目有时间一个个做下来 2009-12-29 15:09 1574人阅读 评论(1) 收藏 举报 图形graphc优化存储游戏 有时间要去做做这些题目,所以从他人空间copy过来了,谢谢那位 ...

  9. hdu 1026:Ignatius and the Princess I(优先队列 + bfs广搜。ps:广搜AC,深搜超时,求助攻!)

    Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (J ...

随机推荐

  1. JS对象的拷贝

    1:对数据进行备份的时候,如果这个数据是基本的数据类型,那么很好办,通过赋值实现复制即可. 赋值与浅拷贝的区别 var obj1 = { 'name' : 'zhangsan', 'age' : '1 ...

  2. Centos6.5升级openssh、OpenSSL和wget

    1.OpenSSL 1.1.查看版本 使用如下命令查看版本: openssl version 1.2.安装gcc依赖 yum -y install gcc gcc-c++ 1.3.安装配置 ./con ...

  3. Exception in thread "main" java.lang.UnsupportedClassVersionError: org/apache/maven/cli/MavenCli : Unsupported major.minor version 51.0 报错

    此报错经常出现,项目中使用的maven版本为3.2.5版本但是去写自动化脚本又需要去3.5.2版本.经常搞混,需要记录一下: 解决如下: 再次install如下: 验证成功!

  4. Java_并发工具包 java.util.concurrent 用户指南(转)

    译序 本指南根据 Jakob Jenkov 最新博客翻译,请随时关注博客更新:http://tutorials.jenkov.com/java-util-concurrent/index.html.本 ...

  5. vue调用Moment显示时间

    1.下载 Moment 网站: http://momentjs.cn/ 2创建一个vue的文本格式    admin.vue 3.定义给值 代码如下 <template> <div ...

  6. ECMA Script 6_ 类 class

    类 class ES6 提供了更接近传统语言的写法,引入了 Class(类)这个概念,作为对象的模板. 通过 class 关键字,可以定义类 class 新的 class 写法只是让对象原型的写法更加 ...

  7. Oracle 备份与恢复

    在进行生产服务器升级.或更换数据库服务器.搭建测试环境时,需要对生产数据库进行备份以及将来可能的还原. 1.expdp导出 expdp DMS version directory=DATA_PUMP_ ...

  8. Dev TreeList 添加节点图标问题

    1. 在设计界面添加imageCollection控件,在属性页设置图标(可Load from disk,也可从Load from dev gallery) 2. TreeList控件有一个叫做Cus ...

  9. Java程序生成一个Access文件

    package access; import java.io.File;import java.io.IOException;import java.sql.SQLException;import j ...

  10. POJ 1324 Holedox Moving (状压BFS)

    POJ 1324 Holedox Moving (状压BFS) Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 18091 Acc ...