BZOJ 3564 SHOI 2014 信号增幅仪 坐标变换+最小圆覆盖

题目大意：给出平面上的一些点，现在让你用一个长轴与x轴成一定角度的，长轴：短轴已知的椭圆来覆盖所有的坐标，求最小的短轴长度。

思路：很明显，这个椭圆的形状和放置状态已经给出了，但是没有办法求最小拖圆覆盖啊。采用坐标变换，将椭圆变成圆。首先我们先让长轴与x轴平行，将平面上的所有点都旋转这个角度。之后只需要让所有点的x坐标除以长轴：短轴就可以了。剩下的就是最小圆覆盖了。

注：坐标旋转公式：

x‘ = x * cos(a) - y * sin(a)

y‘ = x * sin(a) + y * cos(a)

CODE：

#define _CRT_SECURE_NO_WARNINGS

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define MAX 50010
#define PI (acos(-1.0))
using namespace std;

struct Point{
	double x,y;

	Point(double _,double __):x(_),y(__) {}
	Point() {}
	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];

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

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

inline Point Mid(const Point &p1,const Point &p2)
{
	return Point((p1.x + p2.x) / 2,(p1.y + p2.y) / 2);
}

inline Point Change(const Point &v)
{
	return Point(-v.y,v.x);
}

inline void Rotate(Point &p,double alpha)
{
	p = Point(p.x * cos(alpha) - p.y * sin(alpha),p.x * sin(alpha) + p.y * cos(alpha));
}

struct Circle{
	Point o;
	double r;

	Circle(const Point &_,double __):o(_),r(__) {}
	bool InCircle(const Point &p) {
		return Calc(o,p) <= r;
	}
};

struct Line{
	Point p,v;

	Line(const Point &_,const Point &__):p(_),v(__) {}
};

inline Point GetIntersection(const Line &l1,const 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 points;
double a,p;

int main()
{
	cin >> points;
	for(int i = 1; i <= points; ++i)
		point[i].Read();
	cin >> a >> p;
	random_shuffle(point + 1,point + points + 1);
	for(int i = 1; i <= points; ++i) {
		Rotate(point[i],(1 - a / 360) * 2 * PI);
		point[i].x /= p;
	}
	Circle now(point[1],.0);
	for(int i = 2; i <= points; ++i)
		if(!now.InCircle(point[i])) {
			now = Circle(point[i],.0);
			for(int j = 1; j < i; ++j)
				if(!now.InCircle(point[j])) {
					now = Circle(Mid(point[i],point[j]),Calc(point[i],point[j]) / 2);
					for(int k = 1; k < j; ++k)
						if(!now.InCircle(point[k])) {
							Line l1(Mid(point[i],point[j]),Change(point[j] - point[i]));
							Line l2(Mid(point[j],point[k]),Change(point[k] - point[j]));
							Point intersection = GetIntersection(l1,l2);
							now = Circle(intersection,Calc(intersection,point[i]));
						}
				}
		}
	cout << fixed << setprecision(3) << now.r << endl;
	return 0;
}