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莫名其妙就AC了……

圆的反演……

神马是反演？

快去恶补奥数……

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-9;
int dcmp(double x){return fabs(x)<eps?0:x<0?-1:1;}  
struct dot
{
    double x,y;
    dot(){}
    dot(double a,double b){x=a;y=b;}
    dot operator +(dot a){return dot(x+a.x,y+a.y);}
    dot operator -(dot a){return dot(x-a.x,y-a.y);}
    dot operator *(double a){return dot(x*a,y*a);}
    double operator *(dot a){return x*a.y-y*a.x;}
    dot operator /(double a){return dot(x/a,y/a);}
    double operator /(dot a){return x*a.x+y*a.y;}
    bool operator ==(dot a){return x==a.x&&y==a.y;}
    void in(){scanf("%lf%lf",&x,&y);}
    void out(){printf("%f %f\n",x,y);}
    dot norv(){return dot(-y,x);}
    dot univ(){double a=mod();return dot(x/a,y/a);}
    dot ro(double a){return dot(x*cos(a)-y*sin(a),x*sin(a)+y*cos(a));}
    double mod(){return sqrt(x*x+y*y);}
    double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}
};
struct cir
{
    dot o;
    double r;
    cir(){}
    cir(dot a,double b){o=a;r=b;}
    void in(){o.in();scanf("%lf",&r);}
};
struct seg
{
    dot s,e;
    seg(){}
    seg(dot a,dot b){s=a;e=b;}
};
cir sivs(dot a,dot b,dot c)  
{  
    dot dir,a1,b1;
    double t,d,w;
    t=fabs((b-a)*(c-a));   
    d=a.dis(b);  
    t/=d;
    w=0.5/t;
    dir=(b-a).norv();  
    a1=c+dir*(w/d);  
    b1=c-dir*(w/d);  
    if(fabs((b-a)*(a1-a))<fabs((b-a)*(b1-a)))  
        return cir(a1,w);
    else  
        return cir(b1,w);  
}  
cir civs(cir a,dot b)  
{  
    cir c;
    double t,x,y,s;  
    t=a.o.dis(b);  
    x=1.0/(t-a.r);  
    y=1.0/(t+a.r);  
    c.r=(x-y)/2.0;  
    s=(x+y)/2.0;  
    c.o=b+(a.o-b)*(s/t);  
    return c;  
}
seg se[2];
void comseg(dot a,double r1,dot b,double r2)  
{   
    double ang;   
    ang=acos((r1-r2)/a.dis(b));  
    se[0].s=a+(b-a).ro(ang).univ()*r1;  
    se[1].s=a+(b-a).ro(-ang).univ()*r1;  
    ang=pi-ang;  
    se[0].e=b+(a-b).ro(-ang).univ()*r2;  
    se[1].e=b+(a-b).ro(ang).univ()*r2;  
}  
int main()
{
    int T,cnt,i;
    cir a,b,a1,b1,ans[2];
    dot c;
    scanf("%d",&T);
    while(T--)
    {
        a.in();
        b.in();
        c.in();
        a1=civs(a,c);
        b1=civs(b,c);
        comseg(a1.o,a1.r,b1.o,b1.r);
        cnt=0;
        for(i=0;i<2;i++)
            if(dcmp((a1.o-se[i].s)*(se[i].e-se[i].s))==dcmp((c-se[i].s)*(se[i].e-se[i].s)))
                if(dcmp((b1.o-se[i].s)*(se[i].e-se[i].s))==dcmp((c-se[i].s)*(se[i].e-se[i].s)))
					ans[cnt++]=sivs(se[i].s,se[i].e,c);
        printf("%d\n",cnt);
        for(i=0;i<cnt;i++)
            printf("%.8f %.8f %.8f\n",ans[i].o.x,ans[i].o.y,ans[i].r);
    }
}

Problem of Apollonius

1
12 10 1 8 10 1 10 10

2
10.00000000 8.50000000 1.50000000
10.00000000 11.50000000 1.50000000

Hint
This problem is special judged.
 

hdu 4773 Problem of Apollonius

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原文地址：http://blog.csdn.net/stl112514/article/details/41116907