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
- python 网络编程---粘包
一.什么是粘包?(只有在TCP中有粘包现象,在UDP中永远不会粘包) 黏包不一定会发生. 如果发生 了:1.可能是在客户端已经粘了 2.客户端没有粘,可能是在服务端粘了. 所谓的粘包问题:主要是是因为 ...
- tf.train.GradientDescentOptimizer 优化器
tf.train.GradientDescentOptimizer(learning_rate, use_locking=False,name='GradientDescent') 参数: learn ...
- SQL数据类型:nchar,char,varchar,nvarchar 的区别和应用场景
概括: char:固定长度,存储ANSI字符,不足的补英文半角空格.CHAR存储定长数据很方便,CHAR字段上的索引效率级高,比如定义CHAR(10),那么不论你存储的数据是否达到了10个字节,都要占 ...
- fork()系统调用的理解
系统调用fork()用于创建一个新进程.我们可以通过下面的代码来理解,最好是能自己敲一遍运行验证. #include<stdio.h> #include<stdlib.h> ...
- Python工业互联网监控项目实战3—websocket to UI
本小节继续演示如何在Django项目中采用早期websocket技术原型来实现把OPC服务端数据实时推送到UI端,让监控页面在另一种技术方式下,实时显示现场设备的工艺数据变化情况.本例我们仍然采用比较 ...
- search(6)- elastic4s-CRUD
如果我们把ES作为某种数据库来使用的话,必须熟练掌握ES的CRUD操作.在这之前先更正一下上篇中关于检查索引是否存在的方法:elastic4s的具体调用如下: //删除索引 val rspExists ...
- Gatling脚本编写技巧篇(一)
一.公共类抽取 熟悉Gatling的同学都知道Gatling脚本的同学都知道,Gatling的脚本包含三大部分: http head配置 Scenario 执行细节 setUp 组装 那么针对三部分我 ...
- [git] github 推送以及冲突的解决,以及一些命令
推送以及冲突的解决:(我的觉得先看完) (正常情况就是把修改的文件 git add 然后git commit 然后推送就行啦): 下面是一些命令 1.查看分支状态(查看所有:当前检出分支的前面会有星号 ...
- python 工具链 虚拟环境和包管理工具 pipenv
Pipenv is a tool that aims to bring the best of all packaging worlds (bundler, composer, npm, cargo, ...
- 基于Neo4j的个性化Pagerank算法文章推荐系统实践
新版的Neo4j图形算法库(algo)中增加了个性化Pagerank的支持,我一直想找个有意思的应用来验证一下此算法效果.最近我看Peter Lofgren的一篇论文<高效个性化Pagerank ...