UVA - 11178 - Morley's Theorem （计算几何~~）

UVA - 11178

Morley's Theorem

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem D
Morley’s Theorem
Input: Standard Input

Output: Standard Output

 Morley’s theorem states that that the lines trisecting the angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle. For example in the figure below the tri-sectors of angles A, B and C has intersected and created an equilateral triangle DEF.

 

Of course the theorem has various generalizations, in particular if all of the tri-sectors are intersected one obtains four other equilateral triangles. But in the original theorem only tri-sectors nearest to BC are allowed to intersect to get point D, tri-sectors nearest to CA are allowed to intersect point E and tri-sectors nearest to AB are intersected to get point F. Trisector like BD and CE are not allowed to intersect. So ultimately we get only one equilateral triangle DEF. Now your task is to find the Cartesian coordinates of D, E and F given the coordinates of A, B, and C.

 

Input

First line of the input file contains an integer N (0<N<5001) which denotes the number of test cases to follow. Each of the next lines contain six integers . This six integers actually indicates that the Cartesian coordinates of point A, B and C are  respectively. You can assume that the area of triangle ABC is not equal to zero,  and the points A, B and C are in counter clockwise order.

 

Output

For each line of input you should produce one line of output. This line contains six floating point numbers  separated by a single space. These six floating-point actually means that the Cartesian coordinates of D, E and F are  respectively. Errors less than   will be accepted.

 

Sample Input   Output for Sample Input

2 1 1 2 2 1 2 0 0 100 0 50 50

1.316987 1.816987 1.183013 1.683013 1.366025 1.633975 56.698730 25.000000 43.301270 25.000000 50.000000 13.397460

Problemsetters: Shahriar Manzoor

Special Thanks: Joachim Wulff

 

Source

Root :: Prominent Problemsetters ::  Shahriar Manzoor

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 4. Geometry :: Geometric Computations in 2D ::  Examples

思路：首先需要计算角ABC的值b，然后把射线BC逆时针旋转b/3，得到直线BD。同理可求得CD，求交点D即可。。同理可以求得E，F。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

struct Point {
	double x, y;
	Point(double x = 0, double y = 0) : x(x) , y(y) { }  //构造函数，方便代码编写 
};

typedef Point Vector;  //从程序实现上，Vector只是Point的别名 

//向量 + 向量 = 向量 ，点 + 向量 = 点
Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
//点 - 点 = 向量
Vector operator - (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); }
//向量 * 数 = 向量 
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
//向量 / 数 = 向量 
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } 

bool operator < (const Point& a, const Point& b) {
	return a.x < b.x || (a.x == b.x && a.y < b.y);
} 

const double eps = 1e-10;
int dcmp(double x) {
	if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b) {
	return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } //求点积 
double Length(Vector A) { return sqrt(Dot(A, A)); }			 //求向量长度 
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }//求向量之间的夹角 

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }//求叉积 
double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }//根据叉积求三角形面积的两倍 

Vector Rotate(Vector A, double rad) {//rad是弧度 
	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;
} 

Point get_point(Point A, Point B, Point C) {
	Vector v1 = C - B;
	double a1 = Angle(A-B, v1);
	v1 = Rotate(v1, a1/3);
	
	Vector v2 = B - C;
	double a2 = Angle(A-C, v2);
	v2 = Rotate(v2, -a2/3);    //负数表示顺时针旋转 

	return GetLineIntersection(B, v1, C, v2); 
}

int main() {
	int T;
	Point A, B, C, D, E, F;
	scanf("%d", &T);
	while(T--) {
		scanf("%lf %lf %lf %lf %lf %lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
		D = get_point(A, B, C);
		E = get_point(B, C, A);
		F = get_point(C, A, B);
		printf("%lf %lf %lf %lf %lf %lf\n", D.x, D.y, E.x, E.y, F.x, F.y);
	} 
	return 0;
}