Uva - 11178 - Morley's Theorem

题意：求Morley定理的3个点的坐标。

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18543

——>>照要求做～

#include <cstdio>
#include <cmath>

using namespace std;

struct Point{
double x;
double y;
Point(double x = 0, double y = 0):x(x), y(y){}
}p[3];

typedef Point Vector;

Vector operator + (Point A, Point B){
return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Point A, Point B){
return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Point A, double p){
return Vector(A.x * p, A.y * p);
}

Vector operator / (Point A, double p){
return Vector(A.x / p, A.y / p);
}

double Dot(Vector A, Vector B){
return A.x * B.x + A.y * B.y;
}

double Cross(Vector A, Vector B){
return A.x * B.y - B.x * A.y;
}

double Length(Vector A){
return sqrt(Dot(A, A));
}

double Angle(Vector A, Vector B){
return acos(Dot(A, B) / Length(A) / Length(B));
}

Vector Rotate(Vector A, double rad){
return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){
Vector u = P - Q;
double t = Cross(w, u) / Cross(v, w);
return P + v * t;
}

void read(){
for(int i = 0; i < 3; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
}

Point getD(Point A, Point B, Point C){
Vector BC = C - B, BA = A - B;
Vector CB = B - C, CA = A - C;
BC = Rotate(BC, Angle(BC, BA) / 3);
CB = Rotate(CB, -Angle(CB, CA) / 3);
return GetLineIntersection(B, BC, C, CB);
}

void solve(){
Point D, E, F;
D = getD(p[0], p[1], p[2]);
E = getD(p[1], p[2], p[0]);
F = getD(p[2], p[0], p[1]);
printf("%.6f %.6f %.6f %.6f %.6f %.6f\n", D.x, D.y, E.x, E.y, F.x, F.y);
}

int main()
{
int N;
scanf("%d", &N);
while(N--){
read();
solve();
}
return 0;
}