学步园

题目大意：在(0,0)到(10000,10000)的正方形区域内有一些有向直线，求它们左侧面积的交。

思路：半平面交模板题。第一次写半平面交，犯了很多错误。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define MAX 40010
#define EPS 1e-10
#define DCMP(a) (fabs(a) < EPS ? true:false)
using namespace std;

struct Point{
	double x,y;
	
	Point(double _ = 0.0,double __ = 0.0):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 *(const double &a)const {
		return Point(x * a,y * a);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}ploy[MAX],p[MAX];
struct Line{
	Point v,p;
	double alpha;
	
	bool operator <(const Line &a)const {
		return alpha < a.alpha;
	}
	Line(Point _,Point __):v(_),p(__) {}
	Line() {}
	void Calc() {
		alpha = atan2(v.y,v.x);
	}
}line[MAX],q[MAX];

int lines,cnt;

void Pretreatment();
inline double Cross(Point p1,Point p2);
int HalfPlaneIntersection();
inline Point GetIntersection(Line l1,Line l2);
inline bool OnLeft(Line l,Point p);

int main()
{
	Pretreatment();
	cin >> lines;
	for(int i = 1;i <= lines; ++i) {
		Line *now = &line[++cnt];
		now->p.Read();
		Point temp;
		temp.Read();
		now->v = temp - now->p;
	}
	for(int i = 1;i <= cnt; ++i)	line[i].Calc();
	sort(line + 1,line + cnt + 1);
	int num = HalfPlaneIntersection();
	double ans = Cross(ploy[num],ploy[1]);
	for(int i = 1;i < num; ++i)
		ans += Cross(ploy[i],ploy[i + 1]);
	cout << fixed << setprecision(1) << fabs(ans) / 2.0 << endl;
	return 0;
}

void Pretreatment()
{
	line[++cnt] = Line(Point(1,0),Point(0,0));
	line[++cnt] = Line(Point(0,1),Point(10000,0));
	line[++cnt] = Line(Point(-1,0),Point(10000,10000));
	line[++cnt] = Line(Point(0,-1),Point(0,10000));	
}

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

inline Point GetIntersection(Line l1,Line l2)
{
	Point u = l1.p - l2.p;
	double t = Cross(l2.v,u) / Cross(l1.v,l2.v);
	return l1.p + l1.v * t;
}

int HalfPlaneIntersection()
{
	int front = 1,tail = 1;
	q[tail] = line[1];
	for(int i = 2;i <= cnt; ++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(q[tail],line[i].p) ? line[i]:q[tail];
		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;
	if(front == tail)	return 0;
	p[tail] = GetIntersection(q[tail],q[front]);
	int re = 0;
	for(int i = front;i <= tail; ++i)
		ploy[++re] = p[i];
	return re;
}

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