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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <vector>
 7 using namespace std;
 8 
 9 const double EPS = 1e-10;        //精度系数
10 const double PI = acos(-1.0);    //π
11 
12 
13 struct Point {
14     double x, y;
15     Point(double x = 0, double y = 0) :x(x), y(y) {}
16 };    //点的定义
17 
18 typedef Point Vector;    //向量的定义
19 
20 Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }    //向量减法
21 
22 int dcmp(double x) {
23     if (fabs(x) < EPS)return 0; else return x < 0 ? -1 : 1;
24 }    //与0的关系
25 
26 double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }    //向量点乘
27 double Length(Vector A) { return sqrt(Dot(A, A)); }    //向量长度
28 
29 struct Circle {
30     Point c;
31     double r;
32     Circle() :c(Point(0, 0)), r(0) {}
33     Circle(Point c, double r = 0) :c(c), r(r) {}
34     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
35 };    //圆
36 
37 Circle CircumscribedCircle(Point p1, Point p2, Point p3) {
38     double Bx = p2.x - p1.x, By = p2.y - p1.y;
39     double Cx = p3.x - p1.x, Cy = p3.y - p1.y;
40     double D = 2 * (Bx*Cy - By*Cx);
41     double cx = (Cy*(Bx*Bx + By*By) - By*(Cx*Cx + Cy*Cy)) / D + p1.x;
42     double cy = (Bx*(Cx*Cx + Cy*Cy) - Cx*(Bx*Bx + By*By)) / D + p1.y;
43     Point p = Point(cx, cy);
44     return Circle(p, Length(p1 - p));
45 }    //三角形外接圆
46 
47 Point A, B, C;
48 
49 void print(double num) {
50     if (dcmp(num) < 0) { printf("- "); num = -num; } else printf("+ ");
51     printf("%.3lf", num);
52 }    //输出
53 
54 int main() {
55     while (~scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y)) {
56         Circle ans = CircumscribedCircle(A, B, C);
57         double a = ans.c.x, b = ans.c.y;
58         double c = -2 * a, d = -2 * b, e = a*a + b*b - ans.r*ans.r;
59         int tmp = dcmp(a);
60         if (tmp == 0)printf("x^2");
61         else { printf("(x "); print(-a); printf(")^2"); }
62         printf(" + ");
63         tmp = dcmp(b);
64         if (tmp == 0)printf("y^2");
65         else { printf("(y "); print(-b); printf(")^2"); }
66         printf(" = %.3lf^2\n", ans.r);
67         printf("x^2 + y^2 "); print(c); printf("x "); print(d); printf("y "); print(e);
68         printf(" = 0\n\n");
69     }
70 }