POJ 2242 The Circumference of the Circle：计算几何

POJ 2242 The Circumference of the Circle：计算几何2014-07-22 csdn博客 synapse7The Circumference of the Circle

http://poj.org/problem？id=2242

Time Limit: 1000MS

Memory Limit: 65536K

Description

To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don"t？

You are given the cartesian coordinates of three non-collinear points in the plane.

Your job is to calculate the circumference of the unique circle that intersects all three points.

Input

The input will contain one or more test cases. Each test case consists of one line containing six real numbers x1,y1, x2,y2,x3,y3, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.

Output

For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.

Sample Input

0.0 -0.5 0.5 0.0 0.0 0.5

0.0 0.0 0.0 1.0 1.0 1.0

5.0 5.0 5.0 7.0 4.0 6.0

0.0 0.0 -1.0 7.0 7.0 7.0

50.0 50.0 50.0 70.0 40.0 60.0

0.0 0.0 10.0 0.0 20.0 1.0

0.0 -500000.0 500000.0 0.0 0.0 500000.0

Sample Output

3.14

4.44

6.28

31.42

62.83

632.24

3141592.65

Source

Ulm Local 1996

由

S=（a*b*sinC）/2

和

c/sinC=外接圆直径d

得

d=（a*b*c）/（2*S）

S由有向面积（行列式）得出

注意：POJ上的C++不支持C99，但G++支持

完整代码：

/*0ms,588KB*/#include<cstdio>#include<cmath>const double PI = acos（-1.0）;inline double area（double x0, double y0, double x1, double y1, double x2, double y2）{return fabs（x0 * y1 + x2 * y0 + x1 * y2 - x2 * y1 - x0 * y2 - x1 * y0）;}int main（void）{double x0, y0, x1, y1, x2, y2;while （~scanf（"%lf%lf%lf%lf%lf%lf", &x0, &y0, &x1, &y1, &x2, &y2））printf（"%.2f
", hypot（x0 - x1, y0 - y1） * hypot（x1 - x2, y1 - y2） * hypot（x0 - x2, y0 - y2） * PI / area（x0, y0, x1, y1, x2, y2））;return 0;}