/* 1059. Exocenter of a Trian 题目大意： 给出三角形ABC三点坐标，如图所示，作正方形ABDE，正方形BCHJ， 正方形CAGF，L为DG中点，M为EJ中点，N为FH中点，直线LA，MB，NC交于同一点O，求点O的坐标。 其实O就是其垂心 思路：1、先通过向量旋转分别得到D、G、F、H，由于求两直线交点只需两条直线即可， 所以EJ不用求。（向量旋转90） 2、求DG中点L和FH中点N，然后求直线LA和NC的交点 3、这里求两直线交点有三种情况 1.LA无斜率 2.NC无斜率 3.都有斜率 */#include<iostream>#include<cmath>#include<stdlib.h>#include<stdio.h>using namespace std;const double PI = 3.1415926535897932384626433832795;const double eps = 1e-8;int dcmp(double x){return x < -eps ? -1 : x > eps ;} double fix(double x){ if(dcmp(x)==0)return 0; return x;} struct Point { double x, y; Point() {} Point(double x0, double y0): x(x0), y(y0) {}}; double operator*(Point p1, Point p2) // 计算叉乘 p1 × p2{ return (p1.x * p2.y - p2.x * p1.y);}Point operator-(Point p1, Point p2){ return Point(p1.x - p2.x, p1.y - p2.y);}Point operator+(Point p1, Point p2){ return Point(p1.x + p2.x, p1.y + p2.y);}Point Rotate(Point p, double angle){ Point result; result.x = p.x * cos(angle) - p.y * sin(angle); result.y = p.x * sin(angle) + p.y * cos(angle); return result; }double Area(Point A, Point B, Point C) //三角形面积{ return ((B-A)*(C-A) / 2.0);}//无斜率 Point withoutSlop(double x, Point p1, Point p2 ){ double k = (p1.y- p2.y)/(p1.x - p2.x); double b = p1.y - k * p1.x; Point result; result.x = x; result.y = x * k + b; return result; } //都有斜率 Point withSlop(Point p1, Point p2, Point p3, Point p4){ double k1 = (p1.y- p2.y)/(p1.x - p2.x); double b1 = p1.y - k1 * p1.x; double k2 = (p3.y- p4.y)/(p3.x - p4.x); double b2 = p3.y - k2 * p3.x; Point result; result.x = (b1-b2)/(k2-k1); result.y = k1 * result.x + b1; return result; } //求直线交点 Point intersection(Point p1,Point p2,Point p3,Point p4){ Point result; //浮点数绝对值为fabs，整数为abs if(fabs(p1.x - p2.x) < eps) result = withoutSlop(p1.x, p3,p4); else if(fabs(p3.x-p4.x) < eps) result = withoutSlop(p3.x, p1,p2); else result = withSlop(p1,p2,p3,p4); return result; } int main() { int T, cas; Point A, B, C; scanf("%d",&T); for(cas = 0; cas < T; cas++) { scanf("%lf%lf",&A.x, &A.y); scanf("%lf%lf",&B.x, &B.y); scanf("%lf%lf",&C.x, &C.y); if(Area(A,B,C) < 0)swap(B,C); Point D, G, F, H, L, N, ans; //如向量AB=(B-A),旋转90后，变成AD=(D-A); 所以D = Rotate(B-A,-PI/2) + A D = Rotate(B - A,-PI/2) + A; G = Rotate(C - A, PI/2) + A; L = (D + G); L.x/=2; L.y/=2; F = Rotate(A - C,-PI/2) + C; H = Rotate(B - C, PI/2) + C; N = (F + H); N.x/=2; N.y/=2; ans = intersection(A, L, C, N); printf("%.4lf %.4lf\n",fix(ans.x), fix(ans.y)); } system("pause"); return 0; }
