poj 3278 catch that cow BFS(基础水)
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 61826 | Accepted: 19329 |
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
Output
Sample Input
5 17
Sample Output
4
Hint
Source
[Submit] [Go Back] [Status] [Discuss]
Home Page Go Back To top
#include<stdio.h> #include<string.h> #include<queue> #include<iostream> #include<algorithm> using namespace std;
转载于:https://www.cnblogs.com/13224ACMer/p/4738717.html
poj 3278 catch that cow BFS(基础水)相关推荐
- POJ 3278 Catch That Cow BFS
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 32071 Accepted: 9866 D ...
- BFS POJ 3278 Catch That Cow
题目传送门 1 /* 2 BFS简单题:考虑x-1,x+1,x*2三种情况,bfs队列练练手 3 */ 4 #include <cstdio> 5 #include <iostrea ...
- POJ 3278 Catch That Cow
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 30924 Accepted: 9536 D ...
- POJ 3278 Catch That Cow(BFS)
题目网址:http://poj.org/problem?id=3278 题目: Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Tot ...
- bfs+dfs分析----poj 3278 Catch That Cow
题目详情 Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 115430 Accepted: ...
- POJ 3287 Catch That Cow(bfs)
看懂意思就很简单了,给你一个数,每次只能加一或者减一或者乘2,问最少几次才能变成另一个数,就是一个简单的bfs过程,看代码吧. AC代码: #include <iostream> ...
- poj 3278 Catch That Cow(广搜)
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 45087 Accepted: 14116 ...
- POJ 3278 Catch That Cow
Catch That Cow Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 35043 Accepted: 10800 ...
- poj 3278 Catch That Cow 广搜
hdu 2717 Catch That Cow,题目链接 Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Jav ...
最新文章
- element表格取消全选_ElementUi 表格取消全选框,用文字表示
- 什么是分镜头剧本?(分镜头剧本是将文字转换成立体视听形象的中间媒介。主要任务是根据解说词和电视文学脚本来设计相应画面,配置音乐音响,把握片子的节奏和风格等。)
- 【转】使 用免费UMS架设Flash流媒体服务
- Spring Boot WebMagic 入库时 mapper注入提示空指针,以及正确的操作
- java 简单的计算器程序,Java 简易计算器程序
- pyflink shell的remote模式下的实验记录
- php获取服务器名称,PHP 获取服务器详细信息
- OpenTelemetry - 云原生下可观测性的新标准
- jquery validate 中文教程【入门到精通】
- 小学生计算机课堂实践的重要性,小学《信息技术》有效课堂教学的实践与研究课题方案...
- AES-128\192\256加密算法及其安全脆弱分析
- Maven 中 com.adobe.blazeds 的配置失效解决方案
- 华为 IPD 集成产品开发流程的缺点和适用局限性
- 支持傲腾技术的服务器主板,G4560可以用傲腾内存么
- 3des解密 mysql_转角处拐弯007
- 希望各位dalao不吝赐教
- JAVA运行内存的设置
- 对象的属性名和属性值
- 基于Protobuf的通讯库--Poppy简介
- sqlserver 中使用sqlcmd 执行*.sql文件
热门文章
- 【 HDU - 1525 】Euclid's Game(较难找规律,玄学博弈,分析必败点必胜点)
- java读取yaml配置文件,snakeyaml读取yaml配置文件
- c语言编写一个函数判断闰年,C语言:实现一个函数判断year是不是闰年
- MySQL命令(二)| 表的增删查改、聚合函数(复合函数)、联合查询
- 《Head First设计模式》第三章笔记 装饰者模式
- C++:28 --- C++内存布局(上)
- ubuntu apache配置负载均衡篇(一)
- 《Python Cookbook 3rd》笔记(2.18):字符串令牌解析
- android安全 报告,Android安全检测报告
- java图形用户登录界面_Java简单登录图形界面