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最新推荐文章于 2019-01-28 11:45:00 发布

aocai2015

最新推荐文章于 2019-01-28 11:45:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/yoooshinow/p/8318630.html

本文介绍了一种求解最小圆覆盖问题的算法，该问题旨在找到一个圆，使其能够覆盖一组给定的点，并且该圆的半径尽可能小。算法首先通过随机化处理输入点集，然后计算这些点的凸包，最后在凸包上应用最小圆覆盖算法来找到最优解。

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题意：$n$个点，求最小圆覆盖，$n \leq 5e5$

这题数据是随机的hhh

我们可以先求出凸包然后对凸包上的点求最小圆覆盖…（不过直接求应该也行？）

反正随便写好像都能过…

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
const int N=500005;
struct Point
{
    double x,y;
    int rnd;
    Point(double x=0,double y=0):x(x),y(y){}
}p[N],s[N];
struct Line
{
    double k,b;
};
inline bool cmp1(Point a,Point b)
{
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
inline bool cmp2(Point a,Point b)
{
    return a.rnd<b.rnd;
}
inline Point operator +(Point a,Point b)
{
    return Point(a.x+b.x,a.y+b.y);
}
inline Point operator -(Point a,Point b)
{
    return Point(a.x-b.x,a.y-b.y);
}

inline Point operator /(Point a,double d)
{
    return Point(a.x/d,a.y/d);
}
inline double cross(Point a,Point b)
{
    return a.x*b.y-a.y*b.x;
}
inline double sqr2(double x)
{
    return x*x;
}
inline double dist(Point a,Point b)
{
    return sqrt(sqr2(a.x-b.x)+sqr2(a.y-b.y));
}
inline Line getLine(double k,Point a)
{
    Line res;res.k=k;
    res.b=a.y-a.x*k;
    return res;
}
inline Point getLineIntersection(Line l1,Line l2)
{
    Point res;
    res.x=(l2.b-l1.b)/(l1.k-l2.k);
    res.y=res.x*l1.k+l1.b;
    return res;
}
inline Point getCircle(Point a,Point b,Point c)
{
    Point p1=(a+b)/2,p2=(a+c)/2;
    double k1=-(b.x-a.x)/(b.y-a.y);
    double k2=-(c.x-a.x)/(c.y-a.y);
    Line l1=getLine(k1,p1),l2=getLine(k2,p2);
    return getLineIntersection(l1,l2); 
}
inline Point minCircle(double &r,int n)
{
    for(register int i=1;i<=n;i++)s[i].rnd=rand(); 
    sort(s+1,s+n+1,cmp2);
    Point o=s[1];r=0;
    for(register int i=2;i<=n;i++)if(r<dist(o,s[i]))
    {
        o=s[i];r=0;
        for(register int j=1;j<i;j++)if(r<dist(o,s[j]))
        {
            o=(s[i]+s[j])/2;
            r=dist(o,s[i]);
            for(register int k=1;k<j;k++)if(r<dist(o,s[k]))
            {
                o=getCircle(s[i],s[j],s[k]);
                r=dist(o,s[i]);
            }
        }
    }
    return o;
}
inline int convexHull(int n)
{
    sort(p+1,p+n+1,cmp1);
    int t=0,k;
    for(register int i=1;i<=n;i++)
    {
        while(t>1&&cross(s[t]-s[t-1],p[i]-s[t-1])<0)t--;
        s[++t]=p[i];
    }
    k=t;
    for(register int i=n-1;i>=1;i--)
    {
        while(t>k&&cross(s[t]-s[t-1],p[i]-s[t-1])<0)t--;
        s[++t]=p[i];
    }
    if(n>1)t--;
    return t;
}
int main()
{
    int n;scanf("%d",&n);
    for(register int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
    int t=convexHull(n);double r;
    Point res=minCircle(r,t);
    printf("%.2lf %.2lf %.2lf",res.x,res.y,r);
    return 0;
}