BZOJ 1199 HNOI2005 汤姆的游戏 计算几何+暴力

题目大意：给定n个图形，每个图形可以是矩形或圆，m次询问某个点在多少个图形内部

将点按横坐标排序

对于每个图形，二分找到x值满足要求的区间，对于区间内每个点暴力

时间复杂度O(n^2) 数据范围25W

果然像hwd说的一样计算几何题数据范围出的这么大就是作死么= =

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 250250
#define EPS 1e-7
using namespace std;
struct Point{
	double x,y;
	friend istream& operator >> (istream &_,Point &p)
	{
		scanf("%lf%lf",&p.x,&p.y);
		return _;
	}
	bool operator < (const Point &p) const
	{
		return x<p.x;
	}
	friend double Distance(const Point &p1,const Point &p2)
	{
		return sqrt( (p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y) );
	}
};
struct _Point:Point{
	int id;
	friend istream& operator >> (istream &_,_Point &p)
	{
		scanf("%lf%lf",&p.x,&p.y);
		return _;
	}
}points[M];
struct Rectangle{
	Point p1,p2;
	friend istream& operator >> (istream &_,Rectangle &r)
	{
		cin>>r.p1>>r.p2;
		return _;
	}
	bool In_Range(const Point &p) const
	{
		return p1.y+EPS<p.y && p.y<p2.y-EPS;
	}
};
struct Circle{
	Point o;
	double r;
	friend istream& operator >> (istream &_,Circle &c)
	{
		cin>>c.o;
		scanf("%lf",&c.r);
		return _;
	}
	bool In_Range(const Point &p) const
	{
		return Distance(o,p) < r-EPS;
	}
};
struct abcd{
	int type;
	union{
	public:
		Rectangle r;
		Circle c;
	};
	friend istream& operator >> (istream &_,abcd &a)
	{
		char p[10];
		scanf("%s",p);
		if(p[0]=='r')
			a.type=1,cin>>a.r;
		else
			a.type=2,cin>>a.c;
		return _;
	}
	bool In_Range(const Point &p)
	{
		if(type==1)
			return r.In_Range(p);
		else
			return c.In_Range(p);
	}
}a[M];
int n,m,ans[M];
int main()
{
	int i;
	cin>>n>>m;
	for(i=1;i<=n;i++)
		cin>>a[i];
	for(i=1;i<=m;i++)
		cin>>points[i],points[i].id=i;
	sort(points+1,points+m+1);
	for(i=1;i<=n;i++)
	{
		Point l,r;
		if(a[i].type==1)
			l.x=a[i].r.p1.x+EPS,
			r.x=a[i].r.p2.x-EPS;
		else
			l.x=a[i].c.o.x-a[i].c.r+EPS,
			r.x=a[i].c.o.x+a[i].c.r-EPS;
		_Point *_l=lower_bound(points+1,points+m+1,l);
		_Point *_r=lower_bound(points+1,points+m+1,r);
		for(;_l!=_r;_l++)
			if(a[i].In_Range(*_l))
				++ans[_l->id];
	}
	for(i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}