Triangle

2019icpc Nanjing K
比赛的时候怎么没想起来这么简单的做法啊

#include <bits/stdc++.h>
#pragma GCC optimize(3 , "Ofast" , "inline")
using namespace std;
const double eps = 1e-7 ;
struct node
{
   double x , y ;
   bool use ;
}a[10] , b , e , c , d;
double get(node a , node b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)) ;
}
double solve(node a , node b , node c , node d)
{
    return (a.x - b.x) * (c.y - d.y) - (a.y - b.y) * (c.x - d.x) ;
}
double ask(node a , node b , node c , node d)
{
    return (a.x - b.x) * (c.x - d.x) + (a.y - b.y) * (c.y - d.y) ;
}
void calc(node c , node e , double x)
{
//	printf("%.12lf %.12lf %.12lf %.12lf %.12lf
" , c.x , c.y , e.x , e.y , x) ;
    node ce = {e.x - c.x , e.y - c.y , false} ;
    double t = get(c , e) ;
    ce.x = ce.x * x / t , ce.y = ce.y * x / t ;
    printf("%.12lf %.12lf
" , ce.x + c.x , ce.y + c.y) ;
}
int main()
{
    int T ;
    scanf("%d" , &T) ;
    while(T --)
    {
        int f = 0 ;
        for(int i = 1; i <= 3 ;i ++)
             scanf("%lf%lf" , &a[i].x , &a[i].y) ;
        scanf("%lf%lf" , &b.x , &b.y) ;
        for(int i = 1; i <= 3 ;i ++)
        {
          for(int j = 1; j != i && j <= 3 ;j ++)
          {
              double a1 = ask(a[i] , a[j] , a[i] , b) ;
              double a2 = get(a[i] , a[j]) * get(a[i] , b) ;
              double a3 = ask(a[j] , a[i] , a[j] , b) ;
              double a4 = get(a[j] , a[i]) * get(a[j] , b) ;
              if(abs(a1 - a2) <= eps && abs(a3 - a4) <= eps) 
              {
                    e = a[6 - i - j] ;
                    a[i].use = a[j].use = true ;
                    f = 1; 
                    break ;
              }
          }      
            if(f) break ;
        }
        if(!f) puts("-1") ;
        else 
        {
        f = 0 ;
        for(int i = 1; i <= 3 ;i ++)
        {
           for(int j = 1; j != i && j <= 3; j ++)
                if(a[i].use && a[j].use) 
                {
                    if(get(a[i] , b) >= get(a[j] , b))
                    {
                        c = a[i] , d = a[j] ;
                    }
                    else c = a[j] , d = a[i] ;
                    double x = get(c , e) * get(c , d) / (2 * get(c , b)) ;
                     calc(c , e , x) ;
                    f = 1 ;
                    
                    break ;
                }
            if(f) break ;
        }
		}
        
    }
    return 0 ;
}