poj 2242（水题，数学题）

#include<iostream>
#include<cstdio>
using namespace std;
#define PI 3.141592653589793
#include<cmath>
int main(){
    double x1,y1,x2,y2,x3,y3;
    double k1,k2,ansx,ansy,ans;
    while(scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3)==6){
        double a = x1-x2;
        double b = y1-y2;
        double c = x1-x3;
        double d = y1-y3;
        if(a==0){//直线1平行于y轴，中垂线1平行于x轴且过中点 
            k1 = 0; 
            ansy = (y1+y2)/2;
            if(d==0)//直线2平行于x轴，中垂线2平行于y轴且过中点 
                ansx = (x1+x3)/2;
            else{
                k2 = -(x3-x1)/(y3-y1);
//                ansx =    ((y2-y3)-k2*(x1+x3))/(2*k2);
                ansx = (x1+x3)/2+(y2-y3)/2/k2;
            }
        }
        else if(b==0){//直线1平行于x轴，中垂线平行于y轴且过中点
            ansx = (x1+x2)/2;
            if(c==0)//直线2平行于y轴，中垂线2平行于x轴且过中点 
                ansy = (y1+y3)/2;
            else{
                k2 = -(x3-x1)/(y3-y1);
                ansy = (y1+y3)/2+k2*(x2-x3)/2;
            }
        }
        else{
            k1 = -(x2-x1)/(y2-y1);
            if(c==0){//直线2平行于y轴，中垂线2平行于x轴且过中点 
                ansy = (y1+y3)/2;
                ansx = (x1+x2)/2+(y3-y2)/2/k1;
            }else if(d==0){
                ansx = (x1+x3)/2;
                ansy = (y1+y2)/2+k1*(x3-x2)/2;
            }
            else{
                k2 = -(x3-x1)/(y3-y1);
                ansx = (k1*(x1+x2)/2-k2*(x1+x3)/2-(y2-y3)/2)/(k1-k2);
                ansy = k1*(ansx-(x1+x2)/2)+(y1+y2)/2;
            }
        }
        ans = 2*PI*(sqrt((x1-ansx)*(x1-ansx)+(y1-ansy)*(y1-ansy)));
        printf("%.2lf\n",ans);
    }
    return 0;
}