BZOJ 1336 Balkan2002 Alien最小圆覆盖

标签：bzoj   计算几何   最小圆覆盖   随机增量法   

题目大意：最小圆覆盖。

思路：再拍一份模板。做法见：http://blog.csdn.net/jiangyuze831/article/details/43950601

CODE：

#define _CRT_SECURE_NO_WARNINGS

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define MAX 100010
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);
}

struct Circle{
	Point o;
	double r;

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

struct Line{
	Point p,v;

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

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;

int main()
{
	cin >> points;
	for(int i = 1; i <= points; ++i)
		point[i].Read();
	random_shuffle(point + 1,point + points + 1);
	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(10) << now.r << '\n' << now.o.x << ' ' << now.o.y << endl;
	return 0;
}

BZOJ 1336 Balkan2002 Alien最小圆覆盖

标签：bzoj   计算几何   最小圆覆盖   随机增量法   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/43950871