ZOJ 1041 Transmitters

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

该题是一个几何的题目，求给定圆心和半径，求那个位置可以覆盖  给定的点最多，  思路就是 先求得在圆内的点，然后遍历圆内的点，以一个点为基，来和其他的点与圆心连线向量作叉积，求在半圆一侧的点的数目，去最大的。

代码如下

：

#include<stdio.h>#include<string.h>int ju(double a,double b,double c,double d) //求距离 {return (a-c)*(a-c)+(b-d)*(b-d);} int cha(double a,double b,double c,double d,double e,double f)  //作叉积 { return (a-e)*(d-f)-(c-e)*(b-f);}int main(){double x0,y0,r,x[200],y[200],s[200];int i,j,k;while(~scanf("%lf%lf%lf",&x0,&y0,&r)){if(r<0) break;r=r*r;scanf("%d",&k);memset(s,0,sizeof(s));for(i=1;i<=k;i++){scanf("%lf%lf",&x[i],&y[i]);if(ju(x[i],y[i],x0,y0)<=r)  //标记在圆内的点 s[i]=1;}int max=0;int q=0;for(i=1;i<=k;i++){if(s[i]){q=0;for(j=1;j<=k;j++){if(s[j]&&cha(x[i],y[i],x[j],y[j],x0,y0)>=0)  //求在圆一侧的数目 q++;}if(q>max)  max=q;}}printf("%d\n",max);}return 0;}