zoj 1450 Minimal Circle 最小覆盖圆

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=450

You are to write a program to find a circle which covers a set of points and has the minimal area. There will be no more than 100 points in one problem.

题意描述：找到一个最小圆能够包含到所有的二维坐标点。

算法分析：最小覆盖圆的做法。

//最小覆盖圆
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
#define exp 1e-10
#define PI 3.141592654
using namespace std;
const int maxn=1000+10;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}
}an[maxn],d;//d:最小覆盖圆的圆心坐标
double r;//最小覆盖圆的半径
typedef Point Vector;
Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); }
int dcmp(double x)
{
    if (fabs(x)<exp) return 0;
    return x<0 ? -1 : 1;
}
double cross(Vector A,Vector B)
{
    return A.x*B.y-B.x*A.y;
}
double dist(Vector A,Vector B)
{
    double x=(A.x-B.x)*(A.x-B.x);
    double y=(A.y-B.y)*(A.y-B.y);
    return sqrt(x+y);
}

void MiniDiscWith2Point(Point p,Point q,int n)
{
    d=(p+q)/2.0;
    r=dist(p,q)/2;
    int k;
    double c1,c2,t1,t2,t3;
    for (k=1 ;k<=n ;k++)
    {
        if (dist(d,an[k])<=r) continue;
        if (dcmp(cross(p-an[k],q-an[k]))!=0)
        {
            c1=(p.x*p.x+p.y*p.y-q.x*q.x-q.y*q.y)/2.0;
            c2=(p.x*p.x+p.y*p.y-an[k].x*an[k].x-an[k].y*an[k].y)/2.0;
            d.x=(c1*(p.y-an[k].y)-c2*(p.y-q.y))/((p.x-q.x)*(p.y-an[k].y)-(p.x-an[k].x)*(p.y-q.y));
            d.y=(c1*(p.x-an[k].x)-c2*(p.x-q.x))/((p.y-q.y)*(p.x-an[k].x)-(p.y-an[k].y)*(p.x-q.x));
            r=dist(d,an[k]);
        }
        else
        {
            t1=dist(p,q);
            t2=dist(q,an[k]);
            t3=dist(p,an[k]);
            if (t1>=t2 && t1>=t3)
            {
                d=(p+q)/2.0;
                r=dist(p,q)/2.0;
            }
            else if (t2>=t1 && t2>=t3)
            {
                d=(an[k]+q)/2.0;
                r=dist(an[k],q)/2.0;
            }
            else
            {
                d=(an[k]+p)/2.0;
                r=dist(an[k],p)/2.0;
            }
        }
    }
}
void MiniDiscWithPoint(Point p,int n)
{
    d=(p+an[1])/2.0;
    r=dist(p,an[1])/2.0;
    int j;
    for (j=2 ;j<=n ;j++)
    {
        if (dist(d,an[j])<=r) continue;
        else
        {
            MiniDiscWith2Point(p,an[j],j-1);
        }
    }
}

int main()
{
    int n;
    while (scanf("%d",&n)!=EOF && n)
    {
        for (int i=1 ;i<=n ;i++)
        {
            scanf("%lf%lf",&an[i].x,&an[i].y);
        }
        if (n==1)
        {
            printf("%lf %lf
",an[1].x,an[1].y);
            continue;
        }
        r=dist(an[1],an[2])/2.0;
        d=(an[1]+an[2])/2.0;
        for (int i=3 ;i<=n ;i++)
        {
            if (dist(d,an[i])<=r) continue;
            else
            MiniDiscWithPoint(an[i],i-1);
        }
        printf("%.2lf %.2lf %.2lf
",d.x,d.y,r);
    }
    return 0;
}