POJ1329 Circle Through Three Points（三角形外接圆）

题目链接：

　　http://poj.org/problem?id=1329

题目描述：

Circle Through Three Points

 

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

 

题目大意：

　　给三个点求出三角形外接圆

　　并输出圆的标准型和一般型

思路：

　　套模板

　　注意输出处理数值为0（输出x^2）

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;

const double EPS = 1e-10;        //精度系数
const double PI = acos(-1.0);    //π


struct Point {
    double x, y;
    Point(double x = 0, double y = 0) :x(x), y(y) {}
};    //点的定义

typedef Point Vector;    //向量的定义

Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }    //向量减法

int dcmp(double x) {
    if (fabs(x) < EPS)return 0; else return x < 0 ? -1 : 1;
}    //与0的关系

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }    //向量点乘
double Length(Vector A) { return sqrt(Dot(A, A)); }    //向量长度

struct Circle {
    Point c;
    double r;
    Circle() :c(Point(0, 0)), r(0) {}
    Circle(Point c, double r = 0) :c(c), r(r) {}
    Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }    //输入极角返回点坐标
};    //圆

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));
}    //三角形外接圆

Point A, B, C;

void print(double num) {
    if (dcmp(num) < 0) { printf("- "); num = -num; } else printf("+ ");
    printf("%.3lf", num);
}    //输出

int main() {
    while (~scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y)) {
        Circle ans = CircumscribedCircle(A, B, C);
        double a = ans.c.x, b = ans.c.y;
        double c = -2 * a, d = -2 * b, e = a*a + b*b - ans.r*ans.r;
        int tmp = dcmp(a);
        if (tmp == 0)printf("x^2");
        else { printf("(x "); print(-a); printf(")^2"); }
        printf(" + ");
        tmp = dcmp(b);
        if (tmp == 0)printf("y^2");
        else { printf("(y "); print(-b); printf(")^2"); }
        printf(" = %.3lf^2\n", ans.r);
        printf("x^2 + y^2 "); print(c); printf("x "); print(d); printf("y "); print(e);
        printf(" = 0\n\n");
    }
}