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

  1. 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 ...

  2. poj 3348 Cow 凸包面积

    Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8122   Accepted: 3674 Description ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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, ...

  7. 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 ...

  8. POJ 3660 Cow Contest

    题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Subm ...

  9. 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

  1. 分享layui的table的一些小技巧，前端分页

    最近一直在折腾报表,期间使用了layui的table做展示(版本号:2.5.5) 起初:以为是查询结果出来后,前端和服务端分页一弄就完事了,参考例子,但是sql写得太长太长了,翻页困难,数据库是老旧的 ...

  2. 8.4 StringBuilder的介绍及用法（String 和StringBuilder区别）

    * StringBuilder:是一个可变的字符串.字符串缓冲区类.** String和StringBuilder的区别:* String的内容是固定的.(方法区的内容)* StringBuilder ...

  3. 超详细Go语言源码目录说明

    开源项目「go home」聚焦Go语言技术栈与面试题,以协助Gopher登上更大的舞台,欢迎go home~ 导读 学习Go语言源码的第一步就是了解先了解它的目录结构,你对它的源码目录了解多少呢?今天 ...

  4. pgsql的使用

    Deepin上面pgsql的启动 service postgresql start 停止 service postgresql stop 查看pgsql的版本 psql --version

  5. python 性能测试

            python中使用的性能测试模块是memory_profiler , 我们使用它里面的profile这个装饰器即可测试出我们的代码的内存使用情况了.   如果没有安装 memory_p ...

  6. Redis之quicklist源码分析

    一.quicklist简介 Redis列表是简单的字符串列表,按照插入顺序排序.你可以添加一个元素到列表的头部(左边)或者尾部(右边). 一个列表最多可以包含 232 - 1 个元素 (4294967 ...

  7. spring中BeanPostProcessor之三：InitDestroyAnnotationBeanPostProcessor（01）

    在<spring中BeanPostProcessor之二:CommonAnnotationBeanPostProcessor(01)>一文中,分析到在调用CommonAnnotationB ...

  8. 知识点二：HTTP超文本文件传输协议

    HTTP超文本传输协议概念: http1.1之前采用非持续链接服务器在建立连接上开销较大,http1.1之后默认采用持续连接,并有超时设置 http协议:超文本文件传输协议,用于传输文本文件,请求的方 ...

  9. sublime text3添加并修改编译系统

    版权声明:本文为CSDN博主「肥宅_Sean」的原创文章,遵循 CC 4.0 BY-SA 版权协议,原文链接 方法工具 -> 编译系统 -> 新建编译系统 按ctrl+s保存.(注意,这里 ...

  10. L - Neko does Maths CodeForces - 1152C 数论(gcd)

    题目大意:输入两个数 a,b,输出一个k使得lcm(a+k,b+k)尽可能的小,如果有多个K,输出最小的. 题解: 假设gcd(a+k,b+k)=z; 那么(a+k)%z=(b+k)%z=0. a%z ...