UVA 11178 Morley's Theorem（几何）

题目地址：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2119

思路：求出角ABC与角ACB，将向量BC逆时针旋转三分之一角ABC得到向量BD，将向量CB顺时针旋转三分之一角ACB得到向量CD，求向量BD与向量CD的交点即为点D。点E与点F同理。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debug
using namespace std;
const double eps=1e-10;

struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y) {}
void read()
{
scanf("%lf%lf",&x,&y);
}
void print()
{
printf("%.6f %.6f",x,y);
}
};

typedef Point Vector;

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

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);
}

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;
}

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));
}

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

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 A,B,C;

int main()
{
#ifdef debu
freopen("in.txt","r",stdin);
#endif // debug
int t;
scanf("%d",&t);
while(t--)
{
A.read();
B.read();
C.read();

double ABC=Angle(A-B,C-B);
double BAC=Angle(B-A,C-A);
double ACB=Angle(A-C,B-C);

Vector BD=Rotate(C-B,ABC/3);
Vector CD=Rotate(B-C,-ACB/3);
Point D=GetLineIntersection(B,BD/Length(BD),C,CD/Length(CD));

Vector AE=Rotate(C-A,-BAC/3);
Vector CE=Rotate(A-C,ACB/3);
Point E=GetLineIntersection(A,AE/Length(AE),C,CE/Length(CE));

Vector AF=Rotate(B-A,BAC/3);
Vector BF=Rotate(A-B,-ABC/3);
Point F=GetLineIntersection(A,AF/Length(AF),B,BF/Length(BF));

D.print();printf(" ");E.print();printf(" ");F.print();printf(" ");printf("\n");
}
return 0;
}