hdu3867 Light and Shadow

#include <stdexcept>
#include <cstdarg>
#include <iostream>
#include <fstream>
#include <exception>
#include <memory>
#include <locale>
#include <sstream>
#include <set>
#include <list>
#include <bitset>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <string>
#include <utility>
#include <cctype>
#include <climits>
#include <cassert>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <deque>
#include <cstdlib>
#include <stack>
#include <iterator>
#include <functional>
#include <complex>
#include <valarray>

//有时得到的结果是答案的*平方*或*两倍*
//要注意

//计算几何注意 eps
//叉积 cross 是三角形面积的两倍

//向量 (x,y) 逆时针旋转 a 角，x'=x cos a -y sin a , y'=x sin a+y cos a
//确保 不是 0 向量

//直线参数方程 分别为 P+t1v , Q+t2w , u=P-Q , 求交点 , 解得
//t1=cross(w,u)/cross(v,w)
//t2=cross(v,u)/cross(v,w)
//确保有唯一个交点

//求点 P 到直线 AB距离

//一般是弧度不是角度

using namespace std;
const double eps=1e-8;
const double PI=acos(-1.0);
struct Point
{
	double x,y;
	int idx;
	Point(double xx,double yy)
	{
		x=xx;
		y=yy;
		idx=0;
	}
	Point()
	{
		x=0;
		y=0;
		idx=0;
	}
};
typedef Point Vector;
Vector operator+(Vector A,Vector B)
{
    return Vector(A.x+B.x,A.y+B.y);
}
Vector operator-(Point A,Point B)
{
    return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(Vector A,double p)
{
    return Vector(A.x*p,A.y*p);
}
Vector operator/(Vector A,double p)
{
    return Vector(A.x/p,A.y/p);
}
Vector operator*(double p,Vector A)
{
    return Vector(A.x*p,A.y*p);
}
Vector operator/(double p,Vector A)
{
    return Vector(A.x/p,A.y/p);
}
bool operator<(const Point& a,const Point& b)
{
    return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps)
    {
        return 0;
    }
    else
    return x<0?-1:1;
}
bool operator==(const Point& a,const Point& b)
{
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Vector a,Vector b)
{
    //计算点积
    return a.x*b.x+a.y*b.y;
}
double Cross(Vector a,Vector b)
{
    //计算叉积
    return a.x*b.y-a.y*b.x;
}
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
    //求解直线P+t*v  Q+t*w 的交点
    //用前确保两条直线有唯一交点
    //当且仅当cross(v,w)!=0
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
int n,num,T,sum;
Point p1[10005],p2[10005],c,z,x;
double now1,now2;
int a[10005],s[10005];
double l1,l2;
struct Node
{
	Point p;
	double ang;
	int flag;
	int id;
}node[20005];
struct pt
{
	int id;
	pt(int aa)
	{
		id=aa;
	}
	bool operator==(const pt& p) const
	{
		return id==p.id;
	}
	bool operator<(const pt& p) const
	{
		if(id==p.id)
			return false;
		z=Point(0,0);
		x=Point(now1,now2);
		x=GetLineIntersection(z,x-z,p1[id],p2[id]-p1[id]);
		l1=x.x*x.x+x.y*x.y;
		x=Point(now1,now2);
		x=GetLineIntersection(z,x-z,p1[p.id],p2[p.id]-p1[p.id]);
		l2=x.x*x.x+x.y*x.y;
		return l1<l2;
	}
};
map<pt,int> tree;
map<pt,int>::iterator iter;
bool cmp1(Node m,Node n)
{
	return m.ang<n.ang;
}
void solve()
{
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%lf%lf",&c.x,&c.y);
		for(int q=0;q<n;q++)
		{
			scanf("%lf%lf%lf%lf",&p1[q].x,&p1[q].y,&p2[q].x,&p2[q].y);
			p1[q]=p1[q]-c;
			p2[q]=p2[q]-c;
		}
		num=0;
		for(int q=0;q<n;q++)
		{
			if(Cross(Point(0,0)-p1[q],p2[q]-p1[q])>0)
			{
				node[num].p=p1[q];
				node[num].id=q;
				node[num].flag=1;
				node[num].ang=atan2(p1[q].y,p1[q].x);
				num++;
				node[num].p=p2[q];
				node[num].id=q;
				node[num].flag=-1;
				node[num].ang=atan2(p2[q].y,p2[q].x);
				num++;
			}
			else
			{
				node[num].p=p1[q];
				node[num].id=q;
				node[num].flag=-1;
				node[num].ang=atan2(p1[q].y,p1[q].x);
				num++;
				node[num].p=p2[q];
				node[num].id=q;
				node[num].flag=1;
				node[num].ang=atan2(p2[q].y,p2[q].x);
				num++;
			}
		}
		sort(node,node+num,cmp1);
		now1=node[num-1].p.x;
		now2=node[num-1].p.y;
		tree.clear();
		for(int q=0;q<n;q++)
		{
			Point f=GetLineIntersection(Point(0,0),Point(1,0),p1[q],p2[q]-p1[q]);
			if(p1[q].y*p2[q].y<0&&f.x<0)
				tree.insert(make_pair(pt(q),0));
		}
		memset(s,0,sizeof(s));
		s[(tree.begin()->first).id]=1;
		for(int q=0;q<num;q++)
		{
			now1=node[q].p.x;
			now2=node[q].p.y;
			if(node[q].flag==-1)
			{
				tree.insert(make_pair(pt(node[q].id),0));
			}
			else
				tree.erase(tree.find(pt(node[q].id)));
			if(!tree.empty())
				s[(tree.begin()->first).id]=1;
		}
		sum=0;
		for(int q=0;q<n;q++)
			if(s[q]>0)
				sum++;
		printf("%d\n",sum);
	}
}
int main()
{
	solve();
	return 0;
}