POJ 2242 The Circumference of the Circle

The Circumference of the Circle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6480		Accepted: 4077

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

数学题，给出平面直角坐标系中三个不共线点的坐标，求出这三点确定的圆的面积
本来还在求公垂线啊，求交点啊，结果在DISCUSS中看到某大神写的算法，发现自己弱爆了……
大神原话如下：
　　先求出3边长,由海伦公式得出面积
　　S=sqrt(p*(p-a)*(p-b)*(p-c))    p=(a+b+c)/2;
　　由三角形面积公式S=1/2*a*b*sin(C)
　　和正弦定理a/sin(A)=b/sin(B)=c/sin(C)=直径
　　可得直径=a*b*c/2/S;   (a,b,c为三边长)
　　所以周长即为直径*PI

AC代码如下~~

#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

const double pi=3.141592653589793;

double length(double x1,double y1,double x2,double y2)
{
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}

int main()
{
    double x1,y1,x2,y2,x3,y3,a,b,c,p,S,D,C;

    while(cin>>x1>>y1>>x2>>y2>>x3>>y3)
    {
        a=length(x1,y1,x2,y2);
        b=length(x2,y2,x3,y3);
        c=length(x3,y3,x1,y1);
        p=(a+b+c)/2;
        S=sqrt(p*(p-a)*(p-b)*(p-c));
        D=a*b*c/2/S;
        C=pi*D;
        printf("%.2f\n",C);
    }

    return 0;
}