poj 2242(已知三点求外接圆周长)

The Circumference of the Circle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8310		Accepted: 4960

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

开始准备用二分去找的...然后忘了外接圆的定义，然后百度，，发现直接有公式
这里是外接圆半径公式

外接圆：

下面是公式推导：

内接圆

　

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double pi =  3.141592653589793;
const double esp = 1e-8;
struct Point{
    double x,y;
}p[3];
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()
{

    while(scanf("%lf%lf%lf%lf%lf%lf",&p[0].x,&p[0].y,&p[1].x,&p[1].y,&p[2].x,&p[2].y)!=EOF){
        double a = sqrt(dis(p[0],p[1]));
        double b = sqrt(dis(p[1],p[2]));
        double c = sqrt(dis(p[0],p[2]));
        double r = a*b*c/sqrt((a+b+c)*(-a+b+c)*(a-b+c)*(a+b-c));
        printf("%.2lf\n",2*pi*r);
    }
    return 0;
}