POJ 1106 Transmitters 计算几何

题目大意：给出平面内一些点，给一个圆心坐标和圆的半径，求在这个圆心的半圆最多能覆盖多少个点。

思路：暴力枚举就可以，数据量不是很大。先算出那些点有可能成为被覆盖的点，然后枚举所有的点与圆心连线成为圆的半径，用叉积验证是否在圆内即可。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1010
#define EPS 1e-4
using namespace std;

struct Point{
	double x,y;

	Point(double _x,double _y):x(_x),y(_y) {}
	Point() {}
	Point operator -(const Point &a) {
		return Point(x - a.x,y - a.y);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}point[MAX],center;
struct Line{
	Point p1,p2;

	Line(Point _p1,Point _p2):p1(_p1),p2(_p2) {}
	Line() {}
	bool Judge(Point p) {
		Point a = p - p1,b = p - p2;
		return a.x * b.y - b.x * a.y <= 0;
	}
}l;

int points;
bool in_range[MAX];

inline void Initialize();
inline double Calc(Point p1,Point p2);

int main()
{
	double x,y,r;
	while(scanf("%lf%lf",&x,&y) != EOF) {
		Initialize();
		center = Point(x,y);
		scanf("%lf",&r);
		if(r < 0)	continue;
		scanf("%d",&points);
		for(int i = 1;i <= points; ++i) {
			point[i].Read();
			if(Calc(point[i],center) - r <= EPS)
				in_range[i] = true;
		}
		int ans = 0;
		for(int cnt,i = 1;i <= points; ++i) {
			l = Line(center,point[i]);
			cnt = 0;
			for(int j = 1;j <= points; ++j)
				if(in_range[j])
					cnt += l.Judge(point[j]);
			ans = max(ans,cnt);
		}
		printf("%d\n",ans);
	}
	return 0;
}

inline double Calc(Point p1,Point p2)
{
	return sqrt((p1.x - p2.x) * (p1.x - p2.x) + 
				(p1.y - p2.y) * (p1.y - p2.y));
}

inline void Initialize()
{
	memset(in_range,false,sizeof(in_range));
}