SGU414 Orthogonal Circles

来源：互联网发布：软件许可编辑：程序博客网时间：2024/06/04 23:32

题意：给你n个圆，让你找出公共的正交圆。圆的正交就是两圆的圆心分别和交点连线，夹角为90度。
我们可以设出要找的圆，圆心为O，然后我们根据直角三角形的规则列出方程式，最后我们可以得知要找的圆的圆心在两圆连线的某条垂线上，假设两圆的圆心分别为a，b，过圆心O做垂线，在ab上垂足为m，可以得到am^2-bm^2=r1^2-r2^2(为定值)，所以满足条件的圆的圆心在Om那条垂线上。后面就可以根据n个圆解题啦。注意细节。

#include<stdio.h>#include<algorithm>#include<iostream>#include<math.h>#define sqr(x) ((x)*(x))using namespace std;const int maxn = 100000+10;const double eps = 1e-4;const double pi = acos(-1.0);double dcmp(double x){    if(fabs(x)<eps)return 0;    return x>0?1:-1;}double add(double a,double b){//    if(fabs(a+b)<eps*(fabs(a)+fabs(b)))//        return 0;    return a+b;}struct P{    double x,y;    P(double x=0,double y=0):x(x),y(y)    {}    P operator +(P p)    {        return P(add(x,p.x),add(y,p.y));    }    P operator -(P p)    {        return P(add(x,-p.x),add(y,-p.y));    }    P operator *(double d)    {        return P(x*d,y*d);    }    double dot(P p)    {        return add(x*p.x,y*p.y);    }    double det(P p)    {        return add(x*p.y,-y*p.x);    }    bool operator <(const P &a)const    {        return x<a.x||(x==a.x&&y<a.y);    }    bool operator==(const P& a)const    {        return dcmp(x-a.x)==0&&dcmp(y-a.y)==0;    }} node[maxn];struct Circle{    P c;    double r;    Circle() {}    Circle(P c,double r=0):c(c),r(r) {}    P point(double a)    {        return P(c.x+cos(a)*r,c.y+sin(a)*r);    }    bool operator<(const Circle &a)const    {        if(c==a.c)return r<a.r;        else if(dcmp(c.x-a.c.x)==0)return c.y<a.c.y;        return c.x<a.c.x;    }} circle[maxn];struct Line{    P s,e;    Line() {}    Line(P s,P e):s(s),e(e) {}    P point(double a)    {        return s+(e-s)*a;    }} line[maxn];P Rotate(P a,double rad){    return P(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));}P intersection(P p1,P p2,P q1,P q2){    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));}double dist(P a,P b){    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}int main(){    int n;    while(~scanf("%d",&n))    {        int bj=1;        for(int i=0; i<n; i++)        {            scanf("%lf%lf%lf",&circle[i].c.x,&circle[i].c.y,&circle[i].r);        }        if(n==1)        {            printf("-2\n");            continue;        }        sort(circle,circle+n);        int cnt=0;        int yes=0;        for(int i=0; i<n-1; i++)        {            if(circle[i].c==circle[i+1].c&& dcmp(circle[i].r-circle[i+1].r)==0)                continue;            else if(circle[i].c==circle[i+1].c&&dcmp(circle[i].r-circle[i+1].r)!=0)            {                yes=1;                break;            }            else            {                Line AB(circle[i].c,circle[i+1].c);                double dis = dist(circle[i].c,circle[i+1].c);//                printf("%lf %lf %lf\n",circle[i].c.x,circle[i].c.y,dis);                double x=(dis*dis+circle[i].r*circle[i].r-circle[i+1].r*circle[i+1].r)/(2*dis*dis);                P p=AB.point(x);                P temp = circle[i].c-circle[i+1].c;                P q = P(-temp.y*100,temp.x*100)+p;                line[cnt++]=Line(p,q);//                printf("%lf %lf %lf %lf\n",p.x,p.y,q.x,q.y);            }        }        int cnt1=0;        int ok=1;        if(yes)         {            printf("-1\n");            continue;         }        else        {            int zh=0;            for(int i=0; i<cnt-1; i++)            {                if(dcmp((line[i].s-line[i].e).det(line[i+1].s-line[i+1].e))==0)                {                    if(dcmp((line[i].s-line[i+1].s).det(line[i].s-line[i+1].e))==0&&dcmp((line[i].e-line[i+1].s).det(line[i].e-line[i+1].e))==0)                        continue;                    zh=1;                    break;                }                node[cnt1++]=intersection(line[i].s,line[i].e,line[i+1].s,line[i+1].e);            }            if(zh)            {                printf("-1\n");                continue;            }            if(cnt1==0)            {                printf("-2\n");                continue;            }            else            {                int bj0=1;                for(int i=0; i<cnt1-1; i++)                {                    if(node[i]==node[i+1])                    {                        continue;                    }                    bj0=0;                    break;                }                int bj1=1;                if(!bj0)                    printf("-1\n");                else                {//                    printf("%lf %lf\n",node[0].x,node[0].y);                    for(int i=0; i<n; i++)                        if(dcmp(dist(circle[i].c,node[0])-circle[i].r)<=0)                        {                            bj1=0;                            break;                        }                    if(!bj1)                        printf("-1\n");                    else printf("%.10f %.10f %.10f\n",node[0].x,node[0].y,sqrt(sqr(dist(circle[0].c,node[0]))-circle[0].r*circle[0].r));                }            }        }    }    return 0;}/*20 0 510 0 5*/

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