poj 1329 Circle Through Three Points

这个题是求外心的问题；

我们可以用直线相交做，也可以用向量做；

直线相交：

View Code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

using namespace std;

class Point

{

public:

      double x,y;    

};

double circle_R( Point a, double x, double y )

{

    return sqrt( ( a.x - x )*( a.x - x ) + ( a.y -y )*(a.y - y) );    

} 

char sig(double x){return fabs(x)<1e-6 || x>=0?'-':'+';}

void Solve( Point a, Point b ,Point c )

{

     double PY = a.x - b.x ,PX = -a.y + b.y;

     double QY = b.x - c.x ,QX = -b.y + c.y;

     Point p1,p2;

     p1.x = ( a.x + b.x )/2;p1.y = ( a.y + b.y )/2;

     p2.x = ( b.x + c.x )/2;p2.y = ( b.y + c.y )/2;

     double X = ( QX*( PY*p1.x + PX*p2.y ) - PX*( QX*p1.y + QY*p2.x ) )/( PY*QX - PX*QY );

     double Y = ( PY*( QX*p2.y + QY*p1.x ) - QY*( PY*p2.x + PX*p1.y ) )/( PY*QX - PX*QY ); 

     double R = circle_R( a , X ,Y );

     double t = X*X + Y*Y - R*R;

     printf( "(x %c %.3lf)^2 + (y %c %.3lf)^2 = %.3lf^2\n",sig(X),fabs(X),sig(Y),fabs(Y),R );

     printf( "x^2 + y^2 %c %.3lfx %c %.3lfy %c %.3lf = 0\n\n",sig(X),fabs(2*X),sig(Y),fabs(2*Y),sig(-t),fabs(t) );

}

int main(  )

{ 

    Point a,b,c;

    while( scanf( "%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y )==6 )

    {

      Solve( a ,b ,c );

    }

    //system( "pause" );

    return 0;

}

向量相交：

G是△ABC中一点，那么G是△ABC的外心的充要条件：
(GA+GB)·AB=(GB+GC)·BC=(GC+GA)·CA=0

View Code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

using namespace std;

class Point

{

public:

      double x,y;    

};

double circle_R( Point a, double x, double y )

{

    return sqrt( ( a.x - x )*( a.x - x ) + ( a.y -y )*(a.y - y) );    

} 

int dcmp( double x )

{

   if( fabs( x ) < 1.0e-6 )    return 0;

   if( x < 0 ) return -1;

   return 1;

}

char sig(double x){return fabs(x)<1e-6 || x>=0?'-':'+';}

void Solve( Point a, Point b ,Point c )

{

     double A = a.x + b.x,B = a.y + b.y;

     double C = b.x - a.x,D = b.y - a.y;

     double E = c.x + b.x,F = c.y + b.y;

     double G = c.x - b.x,H = c.y - b.y;

     double I = ( A*C + B*D )/2.0;

     double J = ( E*G + F*H )/2.0; 

     double X = ( I*H - J*D )/( C*H - G*D);

     double Y = ( I*G - J*C )/( D*G - C*H);

     double R = circle_R( a , X ,Y );

     double t = X*X + Y*Y - R*R;

     printf( "(x %c %.3lf)^2 + (y %c %.3lf)^2 = %.3lf^2\n",sig(X),fabs(X),sig(Y),fabs(Y),R );

     printf( "x^2 + y^2 %c %.3lfx %c %.3lfy %c %.3lf = 0\n\n",sig(X),fabs(2*X),sig(Y),fabs(2*Y),sig(-t),fabs(t) );

}

int main(  )

{ 

    Point a,b,c;

    while( scanf( "%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y )==6 )

    {

      Solve( a ,b ,c );

    }

    //system( "pause" );

    return 0;

}