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UVA 11178-Morley's Theorem(计算几何_莫雷定理)

时间:2015-05-16 18:25:36      阅读:124      评论:0      收藏:0      [点我收藏+]

标签:计算几何   莫雷定理   

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.

技术分享

 

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 sixintegers 技术分享. This six integers actually indicates that the Cartesian coordinates of point A, B and C are 技术分享 respectively. You can assume that the area of triangle ABC is not equal to zero, 技术分享 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 技术分享 separated by a single space. These six floating-point actually means that the Cartesian coordinates of D, E and F are 技术分享 respectively. Errors less than  技术分享 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

 


莫雷定理:做三角形ABC每个内角的三等分线,相交成三角形DEF,则DEF为等边三角形。

思路:由于等边三角形的对称性,你知道D点就可以,首先计算∠ABC,然后把射线BC逆时针旋转角度的三分之一,得到直线BD,同理可以得到直线CD,求出交点即可。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double eps=1e-10;
const double pi= acos(-1.0);
struct Point {
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
};
typedef Point Vector;

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

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) { //求点积
    return A.x*B.x+A.y*B.y;
}

double Length(Vector A) { //利用点积求向量长度
    return sqrt(Dot(A,A));
}

double Angle(Vector A,Vector B) { //利用点积求夹角
    return acos(Dot(A,B)/Length(A)/Length(B));
}

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

double Area(Point A,Point B,Point C) {
    return Cross(B-A,C-A);
}
Vector Rotate(Vector A,double rad) { //向量的逆时针旋转
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));

}

//调用前确保两条直线相交。当且仅当Cross(v,w)非零。
Point GetLineIntersection(Point P,Vector v, Point Q,Vector w) { //计算两条直线P+tv和Q+tw的交点
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}

Point getD(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=getD(A,B,C);
        E=getD(B,C,A);
        F=getD(C,A,B);
        printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",D.x,D.y,E.x,E.y,F.x,F.y);
    }
    return 0;
}


UVA 11178-Morley's Theorem(计算几何_莫雷定理)

标签:计算几何   莫雷定理   

原文地址:http://blog.csdn.net/u013486414/article/details/45769429

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