Transmitters

最新推荐文章于 2025-09-12 15:58:40 发布

转载最新推荐文章于 2025-09-12 15:58:40 发布 · 217 阅读

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原文链接：http://www.cnblogs.com/lavender913/p/3355758.html

文章标签：

#python #java

本文介绍了一个无线网络中信号覆盖的问题，通过使用一个可以旋转但不能移动的半圆形发射器，来计算其最大能覆盖的点数。文章提供了解决方案，并通过C++代码实现了这一算法。

Transmitters

时间限制(普通/Java):3000MS/10000MS 运行内存限制:65536KByte

描述

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.

输入

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.

输出

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

样例输入

25 25 3.5

25 28

23 27

27 27

24 23

26 23

24 29

26 29

350 200 2.0

350 202

350 199

350 198

348 200

352 200

995 995 10.0

1000 1000

999 998

990 992

1000 999

100 100 -2.5

样例输出

题意：给出圆心坐标及半径，给出n个点的位置，判断半圆在旋转过程中最多能覆盖多少个点，半径输入为负时退出。

解析：先根据点到圆心的距离来判断该点是否在圆里面，在圆外的可剔除；然后根据向量叉积来枚举点与圆心所在的直线（点与圆心所在的直线可将圆分成两部分），判断其他点在此直线的上方或下方，找到点数最多的即为所求答案。

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

double n,m,r;
struct Node
{
    double x;
    double y;
} p[1003];

int main()
{
    int T;
    double x,y,temp;
    while(~scanf("%lf%lf%lf",&n,&m,&r))
    {
        if(r<0) break;
        scanf("%d",&T);
        int k=0;
        for(int i=0; i<T; i++)
        {
            scanf("%lf%lf",&x,&y);
            temp = sqrt((x-n)*(x-n)+(y-m)*(y-m));
            if(temp<=r)
            {
                p[k].x = x-n;
                p[k++].y = y-m;
            }
        }
        int ans , Max = 0;
        for(int i=0; i<k; i++)
        {
            ans = 0;
            for(int j=0; j<k; j++)
            {
                if(p[j].x*p[i].y-p[j].y*p[i].x<=0)
                    ans ++;
            }
            if(ans>Max)
                Max = ans;
        }
        printf("%d\n",Max);
    }
    return 0;
}