poj 2926:Requirements(最远曼哈顿距离,入门题)
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 3908 | Accepted: 1318 |
Description
An undergraduate student, realizing that he needs to do research to improve his chances of being accepted to graduate school, decided that it is now time to do some independent research. Of course, he has decided to do research in the most important domain: the requirements he must fulfill to graduate from his undergraduate university. First, he discovered (to his surprise) that he has to fulfill 5 distinct requirements: the general institute requirement, the writing requirement, the science requirement, the foreign-language requirement, and the field-of-specialization requirement. Formally, a requirement is a fixed number of classes that he has to take during his undergraduate years. Thus, for example, the foreign language requirement specifies that the student has to take 4 classes to fulfill this requirement: French I, French II, French III, and French IV. Having analyzed the immense multitude of the classes that need to be taken to fulfill the different requirements, our student became a little depressed about his undergraduate university: there are so many classes to take…
Dejected, the student began studying the requirements of other universities that he might have chosen after high school. He found that, in fact, other universities had exactly the same 5 requirements as his own university. The only difference was that different universities had different number of classes to be satisfied in each of the five requirement.
Still, it appeared that universities have pretty similar requirements (all of them require a lot of classes), so he hypothesized that no two universities are very dissimilar in their requirements. He defined the dissimilarity of two universities Xand Y as |x1 − y1| + |x2 − y2| + |x3 − y3| + |x4 − y4| + |x5 − y5|, where an xi (yi) is the number of classes in the requirement i of university X (Y) multiplied by an appropriate factor that measures hardness of the corresponding requirement at the corresponding university.
Input
The first line of the input file contains an integer N (1 ≤ N ≤ 100 000), the number of considered universities. The following N lines each describe the requirements of a university. A university X is described by the five non-negative real numbers x1 x2 x3 x4 x5.
Output
On a single line, print the dissimilarity value of the two most dissimilar universities. Your answer should be rounded to exactly two decimal places.
Sample Input
3
2 5 6 2 1.5
1.2 3 2 5 4
7 5 3 2 5
Sample Output
12.80
Source
#define inf 1e200
double a[][]; //每一个点的5个坐标值
double GetManhattan(double p[][],int n,int dem)
{
double ans = ,Min,Max;
int i,j,k;
for(i=;i<(<<(dem-1));i++){ //用二进制形式遍历所有可能的运算情况
Min = inf,Max = -inf;
for(j=;j<n;j++){ //遍历每一个点
double sum = ;
for(k=;k<;k++){ //因为是五维的,所以有4个运算符
//提取当前运算符
int t = i & <<k; //1为+,0为-
if(t) sum+=p[j][k];
else sum-=p[j][k];
}
if(sum>Max) Max = sum;
if(sum<Min) Min = sum;
}
if(Max-Min>ans)
ans = Max - Min;
}
return ans;
}
#include<iostream>
#include<stdio.h>
#include<iomanip>
const double MAX=1e100;
const double MIN=-1e100;
using namespace std;
int n;
double fivemaxdistance() //三维点集最大曼哈顿距离
{
double max1=MIN,min1=MAX,max2=MIN,min2=MAX,max3=MIN,min3=MAX,max4=MIN,min4=MAX,
max5=MIN,min5=MAX,
max6=MIN,min6=MAX,
max7=MIN,min7=MAX,
max8=MIN,min8=MAX,
max9=MIN,min9=MAX,
max10=MIN,min10=MAX,
max11=MIN,min11=MAX,
max12=MIN,min12=MAX,
max13=MIN,min13=MAX,
max14=MIN,min14=MAX,
max15=MIN,min15=MAX,
max16=MIN,min16=MAX,x,y,z,k,l,ans;
for(int i=;i<=n;++i)
{
scanf("%lf%lf%lf%lf%lf",&x,&y,&z,&k,&l);
if(x+y+z+k+l>max1) max1=x+y+z+k+l;
if(x+y+z+k+l<min1) min1=x+y+z+k+l;
if(x-y+z+k+l>max2) max2=x-y+z+k+l;
if(x-y+z+k+l<min2) min2=x-y+z+k+l;
if(x+y-z+k+l>max3) max3=x+y-z+k+l;
if(x+y-z+k+l<min3) min3=x+y-z+k+l;
if(x+y+z-k+l>max4) max4=x+y+z-k+l;
if(x+y+z-k+l<min4) min4=x+y+z-k+l;
if(x+y+z+k-l>max5) max5=x+y+z+k-l;
if(x+y+z+k-l<min5) min5=x+y+z+k-l;
if(x-y-z+k+l>max6) max6=x-y-z+k+l;
if(x-y-z+k+l<min6) min6=x-y-z+k+l;
if(x-y+z-k+l>max7) max7=x-y+z-k+l;
if(x-y+z-k+l<min7) min7=x-y+z-k+l;
if(x-y+z+k-l>max8) max8=x-y+z+k-l;
if(x-y+z+k-l<min8) min8=x-y+z+k-l;
if(x+y-z-k+l>max9) max9=x+y-z-k+l;
if(x+y-z-k+l<min9) min9=x+y-z-k+l;
if(x+y-z+k-l>max10) max10=x+y-z+k-l;
if(x+y-z+k-l<min10) min10=x+y-z+k-l;
if(x+y+z-k-l>max11) max11=x+y+z-k-l;
if(x+y+z-k-l<min11) min11=x+y+z-k-l;
if(x-y-z-k+l>max12) max12=x-y-z-k+l;
if(x-y-z-k+l<min12) min12=x-y-z-k+l;
if(x+y-z-k-l>max13) max13=x+y-z-k-l;
if(x+y-z-k-l<min13) min13=x+y-z-k-l;
if(x-y+z-k-l>max14) max14=x-y+z-k-l;
if(x-y+z-k-l<min14) min14=x-y+z-k-l;
if(x-y-z+k-l>max15) max15=x-y-z+k-l;
if(x-y-z+k-l<min15) min15=x-y-z+k-l;
if(x-y-z-k-l>max16) max16=x-y-z-k-l;
if(x-y-z-k-l<min16) min16=x-y-z-k-l;
}
ans=max1-min1;
if(max2-min2>ans) ans=max2-min2;
if(max3-min3>ans) ans=max3-min3;
if(max4-min4>ans) ans=max4-min4;
if(max5-min5>ans) ans=max5-min5;
if(max6-min6>ans) ans=max6-min6;
if(max7-min7>ans) ans=max7-min7;
if(max8-min8>ans) ans=max8-min8;
if(max9-min9>ans) ans=max9-min9;
if(max10-min10>ans) ans=max10-min10;
if(max11-min11>ans) ans=max11-min11;
if(max12-min12>ans) ans=max12-min12;
if(max13-min13>ans) ans=max13-min13;
if(max14-min14>ans) ans=max14-min14;
if(max15-min15>ans) ans=max15-min15;
if(max16-min16>ans) ans=max16-min16;
return ans;
}
int main()
{
while(cin>>n)
{ //输入点数
cout<<setiosflags(ios::fixed)<<setprecision();
cout<<fivemaxdistance()<<endl;
}
return ;
}
优化方法(用二进制方法遍历运算符):
#include <iostream>
#include <iomanip>
#include <stdio.h>
using namespace std; #define inf 1e200
double a[][]; //每一个点的5个坐标值
double GetManhattan(double p[][],int n,int dem)
{
double ans = ,Min,Max;
int i,j,k;
for(i=;i<(<<(dem-1));i++){ //用二进制形式遍历所有可能的运算情况
Min = inf,Max = -inf;
for(j=;j<n;j++){ //遍历每一个点
double sum = ;
for(k=;k<;k++){ //因为是五维的,所以有4个运算符
//提取当前运算符
int t = i & <<k; //1为+,0为-
if(t) sum+=p[j][k];
else sum-=p[j][k];
}
if(sum>Max) Max = sum;
if(sum<Min) Min = sum;
}
if(Max-Min>ans)
ans = Max - Min;
}
return ans;
}
int main()
{
int i,j,n;
while(cin>>n){
for(i=;i<n;i++) //输入n个点
for(j=;j<;j++)
scanf("%lf",&a[i][j]);
//计算最短曼哈顿距离
cout<<setiosflags(ios::fixed)<<setprecision();
cout<<GetManhattan(a,n,)<<endl;
}
return ;
}
Freecode : www.cnblogs.com/yym2013
poj 2926:Requirements(最远曼哈顿距离,入门题)的更多相关文章
- POJ-2926 Requirements 最远曼哈顿距离
题目链接:http://poj.org/problem?id=2926 题意:求5维空间的点集中的最远曼哈顿距离.. 降维处理,推荐2009武森<浅谈信息学竞赛中的“0”和“1”>以及&l ...
- hdu 4666:Hyperspace(最远曼哈顿距离 + STL使用)
Hyperspace Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Tota ...
- HDU 4666 Hyperspace (最远曼哈顿距离)
Hyperspace Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Tota ...
- poj 2926 Requirements
点击打开poj 2926 思路: n维空间计算最远的曼哈顿距离 分析: 1 题目给定n个5维的点,要求最远的曼哈顿距离 2 求最远曼哈顿距离,对于一个n维的空间,其中两点的曼哈顿距离为:|x1-x2| ...
- HDU 4666 Hyperspace (2013多校7 1001题 最远曼哈顿距离)
Hyperspace Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Tota ...
- 【POJ 3241】Object Clustering 曼哈顿距离最小生成树
http://poj.org/problem?id=3241 曼哈顿距离最小生成树模板题. 核心思想是把坐标系转3次,以及以横坐标为第一关键字,纵坐标为第二关键字排序后,从后往前扫.扫完一个点就把它插 ...
- [HDU 4666]Hyperspace[最远曼哈顿距离][STL]
题意: 许多 k 维点, 求这些点之间的最远曼哈顿距离. 并且有 q 次操作, 插入一个点或者删除一个点. 每次操作之后均输出结果. 思路: 用"疑似绝对值"的思想, 维护每种状态 ...
- HDU 4666 最远曼哈顿距离
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4666 关于最远曼哈顿距离的介绍: http://blog.csdn.net/taozifish/ar ...
- 2018 Multi-University Training Contest 10 CSGO(HDU - 6435)(最远曼哈顿距离)
有 n 种主武器,m 种副武器.每种武器有一个基础分数k种属性值 X[i] . 选出一种主武器 mw 和一种副武器 sw,使得两种武器的分数和 + 每个属性的差值尽量大.(参考下面的式子) 多维的最远 ...
随机推荐
- tomcat8编码
web工程,本机能跑的代码放到生产环境中后能跑但是得不到预期的结果,十有八九的原因是 编码问题
- SYN攻击TIME_WAIT防御。
#! /bin/bash /bin/netstat -anput | grep TIME_WAIT >time_wait.txt /bin/cat time_wait.txt |awk '{pr ...
- linux wget 命令用法详解(附实例说明)
Linux wget是一个下载文件的工具,它用在命令行下.对于Linux用户是必不可少的工具,尤其对于网络管理员,经常要下载一些软件或从远程服务器恢复备份到本地服务器 Linux wget是一个下 ...
- java工具类包
Hutool 提供丰富的java方法,其maven引用 <dependency> <groupId>com.xiaoleilu</groupId> <arti ...
- Protocol Buffers介绍
基本概念 Protocol Buffers(以下简称PB)是一种独立于语言.独立于开发平台.可扩展的序列化数据结构框架,它常常被用在通信.数据序列化保存等方面. PB是一种敏捷.高效.自动化的用于对数 ...
- Microsoft.ReportViewer.WebForms, Version=10.0.0.0的报错问题,解决方案
未能加载文件或程序集,或者web.config报错! 已解决:直接找到(默认在 路径/Microsoft Visual Studio 8/ReportViewer).把里面的3个DLL传上去就OK了! ...
- Linux下安装配置MongoDB 3.0.x 版本数据库
说明: 操作系统:CentOS 5.X 64位 IP地址:192.168.21.128 实现目的: 安装配置MongoDB数据库 具体操作: 一.关闭SElinux.配置防火墙 1.vi /etc/s ...
- 谷歌、百度、1万ip能赚多少钱?1000IP能够值多少钱呢?
谷歌.百度.1万ip能赚多少钱?1000IP能够值多少钱呢? (2014-04-03 11:50:52) 转载▼ 标签: 广告联盟 百度联盟 谷歌联盟 ip赚钱 很多在人问:谷歌.百度:1 ...
- jQuery Ajax 操作函数
jQuery Ajax 操作函数 jQuery 库拥有完整的 Ajax 兼容套件.其中的函数和方法允许我们在不刷新浏览器的情况下从服务器加载数据. 函数 描述 jQuery.ajax() 执行异步 H ...
- 《ASP.NET1200例》高亮显示ListView中的数据行并自动切换图片
aspx <script type="text/javascript"> var oldColor; function SetNewColor(Source) { ol ...