POJ 1329 三角外接圆

Circle Through Three Points

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3169		Accepted: 1342

Description

Your team is to write a program that, given the Cartesian coordinates of three points on a plane, will find the equation of the circle through them all. The three points will not be on a straight line.


The solution is to be printed as an equation of the form

	(x - h)^2 + (y - k)^2 = r^2				(1)

and an equation of the form

	x^2 + y^2 + cx + dy - e = 0				(2)

Input

Each line of input to your program will contain the x and y coordinates of three points, in the order Ax, Ay, Bx, By, Cx, Cy. These coordinates will be real numbers separated from each other by one or more spaces.

Output

Your program must print the required equations on two lines using the format given in the sample below. Your computed values for h, k, r, c, d, and e in Equations 1 and 2 above are to be printed with three digits after the decimal
point. Plus and minus signs in the equations should be changed as needed to avoid multiple signs before a number. Plus, minus, and equal signs must be separated from the adjacent characters by a single space on each side. No other spaces are to appear in the
equations. Print a single blank line after each equation pair.

Sample Input

7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0

Sample Output

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
x^2 + y^2 - 6.000x + 4.000y - 12.000 = 0

(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2
x^2 + y^2 - 7.842x - 4.895y - 7.895 = 0

给定三个点，求三角形的外接圆，题目非常easy，练一下计算几何的模板代码。输出非常恶心。

代码：

/* ***********************************************
Author :rabbit
Created Time :2014/4/19 23:46:03
File Name :8.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)<eps)return 0;
	return x>0?

1:-1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
Point operator + (Point a,Point b){
	return Point(a.x+b.x,a.y+b.y);
}
Point operator - (Point a,Point b){
	return Point(a.x-b.x,a.y-b.y);
}
Point operator * (Point a,double p){
	return Point(a.x*p,a.y*p);
}
Point operator / (Point a,double p){
	return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
	return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
	return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
	return a.x*b.x+a.y*b.y;
}
double Length(Point a){
	return sqrt(Dot(a,a));
}
struct Circle{
	Point c;
	double r;
	Circle(){}
	Circle(Point c,double r):c(c),r(r){}
	Point point(double a){
		return Point(c.x+cos(a)*r,c.y+sin(a)*r);
	}
};
Circle CircumscribedCircle(Point p1,Point p2,Point p3){
	double Bx=p2.x-p1.x,By=p2.y-p1.y;
	double Cx=p3.x-p1.x,Cy=p3.y-p1.y;
	double D=2*(Bx*Cy-By*Cx);
	double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x;
	double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y;
	Point p=Point(cx,cy);
	return Circle(p,Length(p1-p));
}
void output(double R, Point P0)
{
    double C;
    if(P0.x>0)printf("(x - %.3lf)^2 + ",P0.x);else printf("(x + %.3lf)^2 + ",P0.x*(-1));
    if(P0.y>0)printf("(y - %.3lf)^2 = %.3f^2\n",P0.y,R);else printf("(y + %.3lf)^2 = %.3f^2\n",P0.y*(-1),R);
    printf("x^2 + y^2 ");
    if(P0.x>0)printf("- %.3lfx ",P0.x*2);else  printf("+ %.3lfx ",P0.x*(-2));
    if(P0.y>0)printf("- %.3lfy ",P0.y*2);else printf("+ %.3lfy ",P0.y*(-2));
    C = P0.x*P0.x + P0.y*P0.y - R*R;
    if(C>0)printf("+ %.3lf = 0\n",C);else  printf("- %.3lf = 0\n",C*(-1));
}
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     Point a,b,c;
	 Circle p;
	 while(cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y){
		 p=CircumscribedCircle(a,b,c);
		 output(p.r,p.c);
		 puts("");
	 }
     return 0;
}