Transmitters


Time Limit: 2 Seconds      Memory Limit: 65536 KB


In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don't overlap, or at least that they don't conflict. One way of
accomplishing this is to restrict a transmitter's coverage area. This problem uses a shielded transmitter that only broadcasts in a semicircle.



A transmitter T is located somewhere on a 1,000 square meter grid. It broadcasts in a semicircular area of radius r. The transmitter may be rotated any amount, but not moved. Given N points anywhere on the grid, compute the maximum number of points that can
be simultaneously reached by the transmitter's signal. Figure 1 shows the same data points with two different transmitter rotations.

All input coordinates are integers (0-1000). The radius is a positive real number greater than 0. Points on the boundary of a semicircle are considered within that semicircle. There are 1-150 unique points to examine per transmitter. No points are at the same
location as the transmitter.



Input consists of information for one or more independent transmitter problems. Each problem begins with one line containing the (x,y) coordinates of the transmitter followed by the broadcast radius, r. The next line contains the number of points N on the grid,
followed by N sets of (x,y) coordinates, one set per line. The end of the input is signalled by a line with a negative radius; the (x,y) values will be present but indeterminate. Figures 1 and 2 represent the data in the first two example data sets below,
though they are on different scales. Figures 1a and 2 show transmitter rotations that result in maximal coverage.



For each transmitter, the output contains a single line with the maximum number of points that can be contained in some semicircle.

Example input:



25 25 3.5

7

25 28

23 27

27 27

24 23

26 23

24 29

26 29

350 200 2.0

5

350 202

350 199

350 198

348 200

352 200

995 995 10.0

4

1000 1000

999 998

990 992

1000 999

100 100 -2.5

Example output:



3

4

4


题意:用一个固定圆心坐标和半径的半圆去覆盖点,问最多能覆盖的数量

#include <bits/stdc++.h>
using namespace std;
const int N = 205; int x, y;
double r;
int n; int ax[N],ay[N]; int num1,num2; void fun(int i,int j)
{
int tmp=(ax[i]-x)*(ay[j]-y)-(ax[j]-x)*(ay[i]-y);
//两个向量的叉积。a=(x1,y1) b=(x2,y2)
//c=a X b=x1*y2-x2*y1
//假设c>0。表示向量a逆时针到向量b小于PI
//假设c<0,表示向量a顺时针到向量b小于PI
if(tmp==0)
{
num1++;
num2++;
}
else if(tmp>0)
num1++;
else
num2++;
} int main()
{
while(~scanf("%d%d%lf",&x,&y,&r))
{
if(r<0) break; scanf("%d",&n);
int cnt=0;
int a,b; for(int i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
if((a-x)*(a-x)+(b-y)*(b-y)>r*r) continue;
ax[cnt]=a;ay[cnt++]=b;
} int ans=0; for(int i=0;i<cnt;i++)
{
num1=0;
num2=0; for(int j=0;j<cnt;j++)
{
fun(i,j);
}
ans=max(ans,num1);
ans=max(ans,num2);
} printf("%d\n",ans); } return 0;
}

ZOJ1041 Transmitters的更多相关文章

  1. [ACM_几何] Transmitters (zoj 1041 ,可旋转半圆内的最多点)

    Description In a wireless network with multiple transmitters sending on the same frequencies, it is ...

  2. poj 1106 Transmitters (叉乘的应用)

    http://poj.org/problem?id=1106 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4488   A ...

  3. poj1106 Transmitters

    地址:http://poj.org/problem?id=1106 题目: Transmitters Time Limit: 1000MS   Memory Limit: 10000K Total S ...

  4. Poj 1106 Transmitters

    Poj 1106 Transmitters 传送门 给出一个半圆,可以任意旋转,问这个半圆能够覆盖的最多点数. 我们枚举每一个点作为必然覆盖点,那么使用叉积看极角关系即可判断其余的点是否能够与其存在一 ...

  5. POJ 1106 Transmitters(计算几何)

    题目链接 切计算几何,感觉计算几何的算法还不熟.此题,枚举线段和圆点的直线,平分一个圆 #include <iostream> #include <cstring> #incl ...

  6. ZOJ 1041 Transmitters

    原题链接 题目大意:有一个发射站,覆盖范围是半径一定的一个半圆.在一个1000*1000平方米的地盘里有很多接收站.给定发射站的圆心,求最佳角度时能覆盖接收站的个数. 解法:本质上就是给一个原点和其他 ...

  7. 【解题报告】POJ-1106 Transmitters

    原题地址:http://poj.org/problem?id=1106 题目大意: 给定一些平面的点以及一个圆心和半径,过圆心作一个半圆,求点在半圆中点最多多少个. 解题思路: 首先将给定点中和圆心的 ...

  8. poj 1106 Transmitters (枚举+叉积运用)

    题目链接:http://poj.org/problem?id=1106 算法思路:由于圆心和半径都确定,又是180度,这里枚举过一点的直径,求出这个直径的一个在圆上的端点,就可以用叉积的大于,等于,小 ...

  9. UVA 2290 Transmitters

    题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_ ...

随机推荐

  1. drawable的文件名大写

    drawable的文件名大写导致的R文件消失!!!1

  2. MYSQL创建用户和授权方法(测试mysql5.7)

    一.创建用户:  命令:CREATE USER 'username'@'host' IDENTIFIED BY 'password'; 说明:username - 你将创建的用户名, host - 指 ...

  3. JavaScript入门笔记

    第一章 JavaScript语法 1.1 初识JavaScript 1.3 数据类型 1.4 string和boolean类型 1.5 算数操作符 第二章 JavaScript流程控制语句 2.1 循 ...

  4. 如何卸载系统自带的Microsoft Office

    (1)首先.在C盘删除office文件夹. (2)删除注册表 1)开始菜单-->运行-->regedit进入注册表 (window+r  -->) 2)在注册表里找到HKEY_CUR ...

  5. Android文件操作报open failed: EBUSY (Device or resource busy)

    Android删除文件后重新创建时偶尔出现 open failed: EBUSY (Device or resource busy)错误,该错误是Android系统的一个bug,大概的意思类似于win ...

  6. android中textview单行显示,多余的省略

    <TextView android:id="@+id/music_title" android:layout_width="wrap_content" a ...

  7. ElasticSearch-5.21安装

    环境 操作系统:Centos 6.5 X64 IP地址:192.168.56.100 JDK 环境: # java -version java version "1.8.0_121" ...

  8. CAD从线型文件加载线型记录(com接口)

    主要用到函数说明: _DMxDrawX::LoadLinetypeFromFile 从线型文件加载线型记录,详细说明如下: 参数 说明 BSTR pszLinetypeFile 线型文件名,支持htt ...

  9. Git环境部署

     部署git 服务器环境   系统环境准备 192.168.30.25   master     git   gitlab 192.168.30.26   client      git 关闭防火墙 ...

  10. iptables详解(4):iptables匹配条件总结之一

    所属分类:IPtables  Linux基础 在本博客中,从理论到实践,系统的介绍了iptables,如果你想要从头开始了解iptables,可以查看iptables文章列表,直达链接如下 iptab ...