POJ2242_The Circumference of the Circle(几何/三角形外接圆周长/模板)_g. the circumference of the circle description to -优快云博客

本文链接：https://blog.youkuaiyun.com/juncoder/article/details/26163949

本文介绍了一种计算由三个非共线点确定的圆的周长的方法，不依赖于直径信息。通过输入三个点的坐标，可以精确计算出圆的周长，并输出结果保留两位小数。

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The Circumference of the Circle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7017		Accepted: 4353

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

Source

Ulm Local 1996

解题报告

表示几何模板党。。。

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

#define PI  3.141592653589793
using namespace std;
struct point
{
    double x,y;
}p1,p2,p3;
struct line
{
    point a,b;
};
point intersection(line u,line v)
{
    point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))
    /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}
point circumcenter(point a,point b,point c)
{
    line u,v;
    u.a.x=(a.x+b.x)/2;
    u.a.y=(a.y+b.y)/2;
    u.b.x=u.a.x-a.y+b.y;
    u.b.y=u.a.y+a.x-b.x;
    v.a.x=(a.x+c.x)/2;
    v.a.y=(a.y+c.y)/2;
    v.b.x=v.a.x-a.y+c.y;
    v.b.y=v.a.y+a.x-c.x;
    return intersection(u,v);
}
double dis(point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
int main()
{
    while(cin>>p1.x>>p1.y>>p2.x>>p2.y>>p3.x>>p3.y)
    {
        point c=circumcenter(p1,p2,p3);
//        printf("%.2lf %.2lf\n",c.x,c.y);
        double r=dis(p1,c);
        //printf("%.2lf\n",r);
        printf("%.2lf\n",2*PI*r);
    }
    return 0;
}