poj 2242 The Circumference of the Circle

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

这道题我采用的是正弦定理；

一开始忘记自己算的是边长的平方，让我debug了好久；

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#define pi 3.141592653589793
using namespace std;
struct Point
{
    double x;
    double y;
};
double dis(Point a,Point b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int main()
{
    Point point[3];
    while(scanf("%lf %lf %lf %lf %lf %lf",&point[0].x,&point[0].y,&point[1].x,&point[1].y,&point[2].x,&point[2].y)!=EOF)
    {
        double a,b,c;
        a=dis(point[0],point[1]);
        b=dis(point[1],point[2]);
        c=dis(point[0],point[2]);
        double cosc;
        double R,ans;
        cosc=(a+b-c)/(2*sqrt(a)*sqrt(b));
        //cout<<cosc<<endl;
        //cout<<acos(cosc)<<endl;
        //cout<<sin(acos(cosc))<<endl;
        R=sqrt(c)/(sin(acos(cosc)));
        ans=R*pi;
        printf("%.2lf\n",ans);
    }
    return 0;
}