POJ 3384 Feng Shui 半平面交

标签：poj   半平面交   

题目大意：一个人非常信“Feng Shui”，他要在房间里放两个圆形的地毯。这两个地毯之间可以重叠，但是不能折叠，也不能伸到房间的外面。求这两个地毯能够覆盖的最大范围，并输出这两个地毯的圆心。

思路：我们当然希望这两个圆形的地毯离得尽量的远，这样的话两个圆之间的重叠区域就会越小，总的覆盖区域就越大。那我们就先把每一条边向内推进地毯的半径的距离，然后求一次半平面交，这个求出的半平面的交集就是圆心可以取得地方，然后就暴力求出这其中的最远点对就行了。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 110
#define EPS 1e-10
#define DCMP(a) (fabs(a) < EPS)
using namespace std;

struct Point{
	double x,y;

	Point(double _ = .0,double __ = .0):x(_),y(__) {}
	Point operator +(const Point &a)const {
		return Point(x + a.x,y + a.y);
	}
	Point operator -(const Point &a)const {
		return Point(x - a.x,y - a.y);
	}
	Point operator *(double a)const {
		return Point(x * a,y * a);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}point[MAX],p[MAX],polygen[MAX];
struct Line{
	Point p,v;
	double alpha;

	Line(Point _,Point __):p(_),v(__) {
		alpha = atan2(v.y,v.x);
	}
	Line() {}
	bool operator <(const Line &a)const {
		return alpha < a.alpha;
	}
}line[MAX],q[MAX];

int points,lines;
double adjustment;

inline double Cross(const Point &a,const Point &b)
{
	return a.x * b.y - a.y * b.x;
}

inline double Calc(const Point &a,const Point &b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + 
				(a.y - b.y) * (a.y - b.y));
}

inline bool OnLeft(const Line &l,const Point &p)
{
	return Cross(l.v,p - l.p) >= 0;
}

inline void MakeLine(const Point &a,const Point &b)
{
	Point p = a,v = b - a;
	Point _v(-v.y / Calc(a,b),v.x / Calc(a,b));
	p = _v * adjustment + p;
	line[++lines] = Line(p,v);
}

inline Point GetIntersection(const Line &a,const Line &b)
{
	Point u = a.p - b.p;
	double temp = Cross(b.v,u) / Cross(a.v,b.v);
	return a.p + a.v * temp;
}

int HalfPlaneIntersection()
{
	int front = 1,tail = 1;
	q[tail] = line[1];
	for(int i = 2;i <= lines; ++i) {
		while(front < tail && !OnLeft(line[i],p[tail - 1]))	--tail;
		while(front < tail && !OnLeft(line[i],p[front]))	++front;
		if(DCMP(Cross(line[i].v,q[tail].v)))
			q[tail] = OnLeft(line[i],q[tail].p) ? q[tail]:line[i];
		else	q[++tail] = line[i];
		if(front < tail)	p[tail - 1] = GetIntersection(q[tail],q[tail - 1]);
	}
	while(front < tail && !OnLeft(q[front],p[tail - 1]))	--tail;
	p[tail] = GetIntersection(q[tail],q[front]);
	int re = 0;
	for(int i = front;i <= tail; ++i)
		polygen[++re] = p[i];
	return re;
}

pair<Point,Point> GetFarest(int cnt)
{
	double max_length = -1.0;
	pair<Point,Point> re;
	for(int i = 1;i <= cnt; ++i)
		for(int j = i;j <= cnt; ++j)
			if(Calc(polygen[i],polygen[j]) > max_length) {
				max_length = Calc(polygen[i],polygen[j]);
				re.first = polygen[i];
				re.second = polygen[j];
			}
	return re;
}

int main()
{
	cin >> points >> adjustment;
	for(int i = 1;i <= points; ++i)
		point[i].Read();
	for(int i = points;i > 1; --i)
		MakeLine(point[i],point[i - 1]);
	MakeLine(point[1],point[points]);
	sort(line + 1,line + lines + 1);
	int cnt = HalfPlaneIntersection();
	pair<Point,Point> re = GetFarest(cnt);
	printf("%.6lf %.6lf %.6lf %.6lf\n",re.first.x,re.first.y,re.second.x,re.second.y);
	return 0;
}

POJ 3384 Feng Shui 半平面交

标签：poj   半平面交   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40343815