POJ 2188 Cow Laundry
Cow Laundry
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 1376 Accepted: 886
Description
The cows have erected clothes lines with N (1 <= N <= 1000) wires upon which they can dry their clothes after washing them. Having no opposable thumbs, they have thoroughly botched the job. Consider this clothes line arrangement with four wires:
Wire slot: 1 2 3 4
--------------- <-- the holder of the wires
\ \ / /
\ \/ /
\ /\ /
\ / \ / <-- actual clothes lines
/ \
/ \ / \
/ \/ \
/ /\ \
/ / \ \
--------------- <-- the holder of the wires
Wire slot: 1 2 3 4
The wires cross! This is, of course, unacceptable.
The cows want to unscramble the clothes lines with a minimum of hassle. Even their bovine minds can handle the notion of “swap these two lines”. Since the cows have short arms they are limited to swapping adjacent pairs of wire endpoints on either the top or bottom holder.
In the diagram above, it requires four such swaps in order to get a proper arrangement for the clothes line:
1 2 3 4
------------- <-- the holder of the wires
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
------------- <-- the holder of the wires
1 2 3 4
Help the cows unscramble their clothes lines. Find the smallest number of swaps that will get the clothes line into a proper arrangement.
You are supplied with clothes line descriptions that use integers to describe the current ordering of the clothesline. The lines are uniquely numbered 1…N according to no apparent scheme. You are given the order of the wires as they appear in the N connecting slots of the top wire holder and also the order of the wires as they appear on the bottom wire holder.
Input
Line 1: A single integer: N
Lines 2…N+1: Each line contains two integers in the range 1…N. The first integer is the wire ID of the wire in the top wire holder; the second integer is the wire ID of the wire in the bottom holder. Line 2 describes the wires connected to top slot 1 and bottom slot 1, respectively; line 3 describes the wires connected to top and bottom slot 2, respectively; and so on.
Output
- Line 1: A single integer specifying the minimum number of adjacent swaps required to straighten out the clothes lines.
Sample Input
4
4 1
2 3
1 4
3 2
Sample Output
4
Source
USACO 2003 Fall Orange
找到规律的话就是求有多少逆序对,还是比较好想的,但是要处理的话,这个逆序是相对于开始状态,不是另一端点,代码比较简单,思维题,签到题)
#include<iostream>
#include<map>
using namespace std;
int ob[1005];
int oj[1005];
map<int,int>w;
int main()
{
int n;
w.clear();
cin>>n;
for(int i=0;i<n;i++){
cin>>ob[i]>>oj[i];
w.insert(make_pair(ob[i],i));
}
int ans=0;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
if(w[oj[i]]>w[oj[j]]) ans++;
cout<<ans<<endl;
}
POJ 2188 Cow Laundry的更多相关文章
- POJ 3045 Cow Acrobats (贪心)
POJ 3045 Cow Acrobats 这是个贪心的题目,和网上的很多题解略有不同,我的贪心是从最下层开始,每次找到能使该层的牛的风险最小的方案, 记录风险值,上移一层,继续贪心. 最后从遍历每一 ...
- poj 3348 Cow 凸包面积
Cows Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8122 Accepted: 3674 Description ...
- POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest(图论,传递闭包)
POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest(图论,传递闭包) Description N (1 ≤ N ...
- POJ 3176 Cow Bowling(dp)
POJ 3176 Cow Bowling 题目简化即为从一个三角形数列的顶端沿对角线走到底端,所取得的和最大值 7 * 3 8 * 8 1 0 * 2 7 4 4 * 4 5 2 6 5 该走法即为最 ...
- POJ 2184 Cow Exhibition【01背包+负数(经典)】
POJ-2184 [题意]: 有n头牛,每头牛有自己的聪明值和幽默值,选出几头牛使得选出牛的聪明值总和大于0.幽默值总和大于0,求聪明值和幽默值总和相加最大为多少. [分析]:变种的01背包,可以把幽 ...
- POJ 2375 Cow Ski Area(强连通)
POJ 2375 Cow Ski Area id=2375" target="_blank" style="">题目链接 题意:给定一个滑雪场, ...
- Poj 3613 Cow Relays (图论)
Poj 3613 Cow Relays (图论) 题目大意 给出一个无向图,T条边,给出N,S,E,求S到E经过N条边的最短路径长度 理论上讲就是给了有n条边限制的最短路 solution 最一开始想 ...
- POJ 3660 Cow Contest
题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS Memory Limit: 65536K Total Subm ...
- poj 1985 Cow Marathon
题目连接 http://poj.org/problem?id=1985 Cow Marathon Description After hearing about the epidemic of obe ...
随机推荐
- 计算机网络协议,UDP数据报的分析
一.UDP数据报的特点 1.基本特性 UDP是在IP数据报的基础上增加了复用和分用以及差错检测的功能 UDP的主要特点如下: UDP是无连接的:即发送数据之前不需要建立连接 UDP使用尽最大努力交付, ...
- LInux文件管理篇,权限管理
一: chgrp 改变文件所属用户组 chown 改变文件所有者 注意: 1.使用格式 chgrp/chown user file eg: chgrp lanyue permissi ...
- python3(十三)map reduce
# map()函数接收两个参数,一个是函数,一个是Iterable, # map将传入的函数依次作用到序列的每个元素,并把结果作为新的Iterator返回. def f(x): return x * ...
- 127.0.0.1和localhost区别
- Win安装docker
Windows Docker 安装 win7.win8 系统 win7.win8 等需要利用 docker toolbox 来安装,国内可以使用阿里云的镜像来下载,下载地址:http://mirror ...
- c++ 启发式搜索解决八数码问题
本文对八数码问题 启发式搜索 (C++)做了一点点修改 //fn=gn+hn #include<iostream> #include<queue> #include<st ...
- 如何正确管理HBase的连接,从原理到实战
本文将介绍HBase的客户端连接实现,并说明如何正确管理HBase的连接. 最近在搭建一个HBase的可视化管理平台,搭建完成后发现不管什么查询都很慢,甚至于使用api去listTable都要好几秒. ...
- Salesforce元数据入门指南,管理员必看!
元数据是Salesforce基础架构的核心,是Salesforce中的核心组件或功能.没有元数据,大部分功能都无法实现. 但是,某些Salesforce管理员仍然很难掌握元数据的整个范围,并且无法充分 ...
- ASE课程总结 by 冯晓云
开始的开始,采访往届ASE班的blog:http://www.cnblogs.com/legs/p/4894362.html 和北航软工M1检查:http://www.cnblogs.com/legs ...
- Django开发文档-域用户集成登录
项目概述: 一般在企业中,用户以WINDOWS的域用户统一的管理,所以以Django快速开发的应用,不得不集成AD域登录. 网上一般采用django-python3-ldap的库来做集成登录,但是本方 ...