POJ 2242 The circumference of Circle: computational geometry

The 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 is an easy task-provided your know its diameter. But What if you don ' t?

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

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

Input

The input would contain one or more test cases. Each test case consists of the ' one line ' containing six real numbers x1,y1, X2,y2,x3,y3, representing the ' th ' Ree points. The diameter of the circle determined by the three points'll never exceed a million. The Input is terminated by the end of file.

Output

For each test case, print one line containing one real number telling the circumference of the circle determined by the TH Ree points. The circumference is 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

By

s= (a*b*sinc)/2

And

c/sinc= circumcircle diameter d

Have

D= (a*b*c)/(2*s)

S is derived from the area of the direction (determinant)

Note: C + + on POJ does not support C99, but g++ supports

Complete code:

/*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\n", 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;  
}

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