标签：

增量法最小圆覆盖,简单模版

Buried memory

3
1.00 1.00
2.00 2.00
3.00 3.00
0

2.00 2.00 1.41

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月29日 星期一 15时15分46秒
File Name     :HDOJ3007.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn = 888;
const double eps=1e-8;

struct Point
{
	double x,y;
}pt[maxn];

Point operator+(Point A,Point B) { return (Point){A.x+B.x,A.y+B.y}; }
Point operator-(Point A,Point B) { return (Point){A.x-B.x,A.y-B.y}; }
Point operator*(Point A,double p) { return (Point){A.x*p,A.y*p}; }
Point operator/(Point A,double p) { return (Point){A.x/p,A.y/p}; }

int n;

int dcmp(double x)
{
	if(fabs(x)<eps) return 0;
	if(x>eps) return 1;
	else return -1;
}

double Cross(Point A,Point B) { return A.x*B.y-A.y*B.x; }
double Dot(Point a,Point b) { return a.x*b.x+a.y*b.y; }
double Length(Point a) { return sqrt(Dot(a,a)); }

struct Circle
{
	Point c; double r;
};

Circle CircumscribedCircle(Point p1,Point p2,Point p3)
{
	double Bx=p2.x-p1.x,By=p2.y-p1.y;
	double Cx=p3.x-p1.x,Cy=p3.y-p1.y;
	double D=2*(Bx*Cy-By*Cx);
	double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x;
	double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y;
	Circle c;
	c.c=(Point){cx,cy};
	c.r=Length(p1-c.c);
	return c;
}

void min_cover_circle(Point p[],int n,Circle& c)
{
	c.c=p[0]; c.r=0;
	for(int i=1;i<n;i++)
	{
		if(dcmp(Length(p[i]-c.c)-c.r)>0)
		{
			c.c=p[i]; c.r=0;
			for(int j=0;j<i;j++)
			{
				if(dcmp(Length(p[j]-c.c)-c.r)>0)
				{
					c.c=(Point){(p[i].x+p[j].x)/2,(p[i].y+p[j].y)/2};
					c.r=Length(p[j]-p[i])/2.;
					for(int k=0;k<j;k++)
					{
						if(dcmp(Length(p[k]-c.c)-c.r)>0)
						{
							c=CircumscribedCircle(p[i],p[j],p[k]);
						}
					}
				}
			}
		}
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	while(scanf("%d",&n)!=EOF&&n)
	{
		for(int i=0;i<n;i++) scanf("%lf%lf",&pt[i].x,&pt[i].y);
		Circle c;
		min_cover_circle(pt,n,c);
		printf("%.2lf %.2lf %.2lf\n",c.c.x,c.c.y,c.r);
	}
    
    return 0;
}

HDOJ 3007 Buried memory 增量法最小圆覆盖

标签：

原文地址：http://blog.csdn.net/ck_boss/article/details/42240017