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;}
好想从现在开始抱着你，紧紧地抱着你，一直走到上帝面前。
