学步园

【题目大意】

给定一个凸四边形的三条等长边的中点，求这个四边形的四个顶点坐标

【输入】

第一行一个数字T

接下来T行每行六个数字描述中点的坐标

【输出】

如果存在凸四边形，输出"YES"并在下一行按顺时针或逆时针顺序输出四个点的坐标

如果不存在，输出"NO"并在下一行输出一个空行

首先枚举哪个点是被两条等长边夹住的等长边的中点(xb,yb)，另外两个点我们记为(xa,ya),(xc,yc)

然后设这条边的两个顶点(x1,y1),(x2,y2)

我们有

x1+x2=2*xb

y1+y2=2*yb

(x1-xb)^2+(y1-yb)=(x1-xa)^2+(x1-ya)

(x2-xb)^2+(y2-yb)=(x2-xc)^2+(x2-yc)

解方程

然后可以知道四个顶点的坐标

再判下对边是否相交

再判下是不是退化成三角形或者线段或者点

再判下是不是凸多边形

极其蛋疼

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const double eps=1e-7;
int t,i,j,k;
int x[4],y[4];

double fabs(double x)
{
       if (x>0) return x;else return -x;
}

double cross(double x1,double y1,double x2,double y2,double x3,double y3)
{
       double u1,v1,u2,v2;
       u1=x1-x2;
       v1=y1-y2;
       u2=x3-x2;
       v2=y3-y2;
       return u1*v2-u2*v1;
}

bool meet(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4)
{
     double u1,v1,u2,v2,u3,v3;
     u1=x1-x2;
     v1=y1-y2;
     u2=x3-x2;
     v2=y3-y2;
     u3=x4-x2;
     v3=y4-y2;
     return (u1*v2-u2*v1)*(u1*v3-u3*v1)<eps;
}

bool solve(int a,int b,int c)
{
     if ((x[b]-x[a])*(y[c]-y[a])-(x[c]-x[a])*(y[b]-y[a])==0) return false;
     double a1,a2,b1,b2,c1,c2,x1,x2,y1,y2;
     a1=2*(x[a]-x[b]);
     b1=2*(y[a]-y[b]);
     c1=x[a]*x[a]+y[a]*y[a]-x[b]*x[b]-y[b]*y[b];
     x2=2*x[b]-x[c];
     y2=2*y[b]-y[c];
     a2=2*(x2-x[b]);
     b2=2*(y2-y[b]);
     c2=x2*x2+y2*y2-x[b]*x[b]-y[b]*y[b];
     if (fabs(a1)<eps)
        swap(a1,a2),swap(b1,b2),swap(c1,c2);
     if (fabs(a1)<eps) return false;
     if (fabs(a2)<eps)
     {
        y1=c2/b2;
        x1=(c1-b1*y1)/a1;
        x2=2*x[b]-x1;
        y2=2*y[b]-y1;
     }
     else
     {
         if (fabs(b1/a1-b2/a2)<eps && fabs(c1/a1-c2/a2)<eps) return false;
         y1=(c1-c2*a1/a2)/(b1-b2*a1/a2);
         x1=(c1-b1*y1)/a1;
         x2=2*x[b]-x1;
         y2=2*y[b]-y1;
     }
     double x4=2*x[a]-x1,y4=2*y[a]-y1;
     double x3=2*x[c]-x2,y3=2*y[c]-y2;
     if (meet(x[a],y[a],x4,y4,x[c],y[c],x3,y3)) return false;
     if (
        fabs((x1-x2)*(y3-y2)-(x3-x2)*(y1-y2))<eps ||
        fabs((x2-x3)*(y4-y3)-(x4-x3)*(y2-y3))<eps ||
        fabs((x3-x4)*(y1-y4)-(x1-x4)*(y3-y4))<eps ||
        fabs((x2-x1)*(y4-y1)-(x4-x1)*(y2-y1))<eps 
        ) return false;
     if (meet(x1,y1,x2,y2,x3,y3,x4,y4)) return false;
     if (meet(x2,y2,x3,y3,x1,y1,x4,y4)) return false;
     if (
        cross(x1,y1,x2,y2,x3,y3)*cross(x2,y2,x3,y3,x4,y4)<eps ||
        cross(x2,y2,x3,y3,x4,y4)*cross(x3,y3,x4,y4,x1,y1)<eps ||
        cross(x3,y3,x4,y4,x1,y1)*cross(x4,y4,x1,y1,x2,y2)<eps ||
        cross(x4,y4,x1,y1,x2,y2)*cross(x1,y1,x2,y2,x3,y3)<eps
        ) return false;
     cout << "YES" << endl;
     printf("%.9f %.9f %.9f %.9f %.9f %.9f %.9f %.9f\n",x1,y1,x2,y2,x3,y3,x4,y4);
     return true;
}

int main()
{
    cin >> t;
    while (t--)
    {
          for (i=1;i<=3;i++) cin >> x[i] >> y[i];
          if ((x[1]==x[2] && y[1]==y[2])||(x[1]==x[3] && y[1]==y[3])||(x[2]==x[3] && y[2]==y[3]))
             {cout << "NO" <<endl << endl;continue;}
          if (solve(1,2,3)) continue;
          if (solve(2,1,3)) continue;
          if (solve(2,3,1)) continue;
          cout << "NO" << endl << endl;
    }
    return 0;
}