H-Magic Line_2019 牛客暑期多校训练营(第三场)
题目连接:
https://ac.nowcoder.com/acm/contest/883/H
Description
There are always some problems that seem simple but is difficult to solve.
ZYB got N distinct points on a two-dimensional plane. He wants to draw a magic line so that the points will be divided into two parts, and the number of points in each part is the same. There is also a restriction: this line can not pass through any of the points.
Help him draw this magic line.
Input
There are multiple cases. The first line of the input contains a single integer \(T(1<=T<=1000)\), indicating the number of cases.
For each case, the first line of the input contains a single even integer \(N (2 <= N <= 1000)\), the number of points. The following \(N\) lines each contains two integers xi,yi |(xi,yi)| <= 1000, denoting the x-coordinate and the y-coordinate of the -th point.
It is guaranteed that the sum of N over all cases does not exceed 2*10^5.
Output
For each case, print four integers \(x_1, y_1, x_2, y_2\) in a line, representing a line passing through \((x_1, y_1)\) and$ (x_2, y_2)$. Obviously the output must satisfy .
The absolute value of each coordinate must not exceed \(10^9\). It is guaranteed that at least one solution exists. If there are multiple solutions, print any of them.
Sample Input
1
4
0 1
-1 0
1 0
0 -1
Sample Output
-1 999000000 1 -999000001
Hint
题意
二维平面上有n个整数坐标的点,求出一条直线将平面上的点分为数量相等的两部分,且线上不能有点,输出线上两个点确定该直线
题解:
先在左下角无穷远处取一质数坐标点(x,y) 对该点和n个点进行极角排序,设排序后中点坐标为(a,b)则这两点连线会将点分为数量相等的两部分,接着取左下角关于中点的对称点(a+a-x, b+b-y),再将该点左移动一格变成(2a-x-1, 2b-y)
则(x,y) (2a-x-1, 2b-y)两点确定的直线就可以分割点为两部分,且线上不会有点
代码
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX=100005;
const int INF=999999;
typedef long long ll;int n,top;
struct Node
{ll x,y;
}p[MAX],S[MAX];
ll Cross(Node a,Node b,Node c)
{return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}ll dis(Node a,Node b)
{return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}bool cmp(Node a,Node b)
{ ll flag = Cross(p[1],a,b);if(flag != 0) return flag > 0;return dis(p[1],a) < dis(p[1],b);
}int main()
{int T;scanf("%d", &T);while(T--){int n;scanf("%d", &n);p[1].x = -400000009; p[1].y = -2e3;for(int i=1;i<=n;i++)scanf("%lld%lld",&p[i+1].x,&p[i+1].y);n++;sort(p+2, p+1+n, cmp);int pos = n/2 + 1;ll a = p[pos].x - p[1].x + p[pos].x-1;ll b = p[pos].y - p[1].y + p[pos].y;printf("%lld %lld %lld %lld\n", p[1].x, p[1].y, a, b);}
}
转载于:https://www.cnblogs.com/bpdwn-cnblogs/p/11247943.html
H-Magic Line_2019 牛客暑期多校训练营(第三场)相关推荐
- 2019牛客暑期多校训练营 第三场 I Median
传送门 链接:https://ac.nowcoder.com/acm/contest/883/I 来源:牛客网 JSB has an integer sequence a1,a2,-,ana_1, a ...
- 2020牛客暑期多校训练营(第一场)
文章目录 A B-Suffix Array B Infinite Tree C Domino D Quadratic Form E Counting Spanning Trees F Infinite ...
- 2020牛客暑期多校训练营(第二场)
2020牛客暑期多校训练营(第二场) 最烦英语题 文章目录 A All with Pairs B Boundary C Cover the Tree D Duration E Exclusive OR ...
- E Groundhog Chasing Death(2020牛客暑期多校训练营(第九场))(思维+费马小定理+质因子分解)
E Groundhog Chasing Death(2020牛客暑期多校训练营(第九场))(思维+费马小定理+质因子分解) 链接:https://ac.nowcoder.com/acm/contest ...
- 【2019牛客暑期多校训练营(第二场) - H】Second Large Rectangle(单调栈,全1子矩阵变形)
题干: 链接:https://ac.nowcoder.com/acm/contest/882/H 来源:牛客网 题目描述 Given a N×MN \times MN×M binary matrix. ...
- 【2019牛客暑期多校训练营(第一场) - H】XOR(线性基,期望的线性性)
题干: 链接:https://ac.nowcoder.com/acm/contest/881/H 来源:牛客网 Bobo has a set A of n integers a1,a2,-,ana1, ...
- 2019牛客暑期多校训练营(第九场)H Cutting Bamboos(主席树+二分)
链接:https://ac.nowcoder.com/acm/contest/889/H 来源:牛客网 题目描述 There are n bamboos arranged in a line. The ...
- 2019牛客暑期多校训练营(第九场)A——The power of Fibonacci(循环节+中国剩余定理(互质)||广义BM)
链接:https://ac.nowcoder.com/acm/contest/889/A 来源:牛客网 时间限制:C/C++ 2秒,其他语言4秒 空间限制:C/C++ 262144K,其他语言5242 ...
- 2020牛客暑期多校训练营(第一场)A B-Suffix Array(后缀数组,思维)
链接:https://ac.nowcoder.com/acm/contest/5666/A 来源:牛客网 题目描述 The BBB-function B(t1t2-tk)=b1b2-bkB(t_1 t ...
- 题解 | Popping Balloons-2019牛客暑期多校训练营第十场F题
题目来源于牛客竞赛:https://ac.nowcoder.com/acm/contest/discuss 题目描述: 输入描述: 输出描述: 示例1: 示例2: 题解: 代码: #include&l ...
最新文章
- 解密jQuery事件核心 - 绑定设计(一)
- 两篇关于MCU的嵌入式应用的文章【ZZ】
- 类与接口(三)java中的接口与嵌套接口
- 使用Seq2Seq+attention实现简单的Chatbot
- k8s核心技术-Ingress(概述)---K8S_Google工作笔记0041
- Asp.net中模仿Winform的MessageBox
- [CareerCup] 12.6 Test an ATM 测试一个自动取款机
- ubuntu安装公式编辑器mathtype, wine中文乱码,ubuntu中文字体
- Chrome插件——一键保存网页为PDF1.0发布
- android修改shell串口号,[Note] 2021-01-15 Android shell/串口中使用 wpa_cli 连接Wi-Fi
- sed解析url的域名
- 关于高通8953开机需要按pwrkey很长时间的问题
- OV7725摄像头之OV7725芯片
- robo3T-操作MongoDB数据库常用命令
- 第二证券|钠电池三种技术路线谁更将率先取代锂电池?
- Android Studio使用技巧系列教程(四)
- python+django大学生专业社团信息管理系统
- 老调重弹之ffmpeg解码音频
- stats | 广义线性模型(三)——二元Logistic模型和Probit模型
- (附源码)计算机毕业设计ssm二手图书回收销售网站
热门文章
- spring-boot-maven-plugin多模块install问题解决办法
- 如何更好使用 ng-zorro-antd 图标
- 如何提高 Java 中锁的性能
- Java内嵌Groovy脚本引擎进行业务规则剥离(一)
- 魅族2016Java互联网方向其中一道笔试题--青蛙跳台阶问题
- 《梦断代码》读书笔记——第3、4、5章
- linux下find用法 find -name *.so -exec ll {} \;
- 蓝桥杯 ADV-149 算法提高 特殊的质数肋骨
- c#与马扎克通讯_马扎克伺服报警
- 获取对象的key_玩转 SpringBoot2.x 之缓存对象