ZOJ1041 Transmitters
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
#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
- [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 ...
- poj 1106 Transmitters (叉乘的应用)
http://poj.org/problem?id=1106 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4488 A ...
- poj1106 Transmitters
地址:http://poj.org/problem?id=1106 题目: Transmitters Time Limit: 1000MS Memory Limit: 10000K Total S ...
- 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 ...
- 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 ...
- 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 ...
- 【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 ...
- 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 ...
- 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
- Java——Spring注解
Spring常用注解使用注解来构造IoC容器用注解来向Spring容器注册Bean.需要在applicationContext.xml中注册<context:component-scan bas ...
- Java&Xml教程(一)简介
XML是广泛用于数据传输和存储的技术.Java语言提供个各种各样的API来解析XML,例如DOM.SAX.StAX.JAXB.也还有一些其他的API用于解析XML,例如JDOM.本教程的目的是探索使用 ...
- 实现PC延迟执行函数
头文件内容: #pragma once typedef function<void ()> DelayClickHandler; typedef void (*pDelayFun)(); ...
- Win32子窗口的创建
本文主要是在一个主窗口下创建一个子窗口.主窗口有一个菜单,菜单下只有设置一个选项,点击设置选项,弹出设置界面,点击设置界面关闭则关闭.我在开发的时候遇到两个问题,第一就是一点设置关闭就整个应用都关了, ...
- 02--Java Socket编程--IO方式
一.基础知识 1. TCP状态转换知识,可参考: http://www.cnblogs.com/qlee/archive/2011/07/12/2104089.html 2. 数据传输 3. TCP/ ...
- jQuery——入口函数
中文网 http://www.css88.com/jqapi-1.9/ 版本兼容问题 版本一:1.x版本,兼容IE678 版本二:2.x版本,不兼容IE678 入口函数区别 <script> ...
- JS——stye属性
1.样式少的时候使用 this.parentNode.style.backgroundColor="yellow"; 2.style是对象 console.log(box.styl ...
- (转)OL2中设置鼠标的样式
http://blog.csdn.net/gisshixisheng/article/details/49496289 概述: 在OL2中,鼠标默认是箭头,地图移动时,鼠标样式是移动样式:很多时候,为 ...
- CAD在网页中如何设置实体闪烁?
主要用到函数说明: MxDrawXCustomFunction::Mx_TwinkeEnt 闪烁实体.详细说明如下: 参数 说明 McDbObjectId id 被闪烁的实体对象id LONG lCo ...
- Centos7自动式脚本搭建jumpserver
JumpServer脚本 这里需要安装阿里的yum源和epel源并解压: epel源地址https://mirrors.tuna.tsinghua.edu.cn/epel// 安装阿里互联网yum仓库 ...