[Uva11178]Morley's Theorem（计算几何）

Description

题目链接

Solution

计算几何入门题

只要求出三角形DEF的一个点就能推出其他两个点

把一条边往内旋转a/3度得到一条射线，再做一条交点就是了

Code

#include <cstdio>

#include <algorithm>

#include <cmath>

#define db double

using namespace std;

struct Po{

	db x,y;

	Po(db x=0,db y=0):x(x),y(y){}

};

typedef Po Ve;

Ve operator - (Po A,Po B){return Ve(A.x-B.x,A.y-B.y);}

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

Ve operator * (Ve A,db p){return Ve(A.x*p,A.y*p);}

Ve operator / (Ve A,db p){return Ve(A.x/p,A.y/p);}

Po read_p(){

	Po res;

	scanf("%lf%lf",&res.x,&res.y);

	return res;

}

db Dot(Ve A,Ve B){return A.x*B.x+A.y*B.y;}

db Len(Ve A){return sqrt(Dot(A,A));}

db Angle(Ve A,Ve B){return acos(Dot(A,B)/Len(A)/Len(B));}

Ve Rotate(Ve A,db rad){

	return Ve(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));

}

db Corss(Ve A,Ve B){return A.x*B.y-A.y*B.x;}

Po GetIntersection(Po P,Ve v,Po Q,Ve w){

	Ve u=P-Q;

	db t=Corss(w,u)/Corss(v,w);

	return P+v*t;

}

Po getD(Po A,Po B,Po C){

	Ve v1=C-B;

	db a1=Angle(A-B,v1);

	v1=Rotate(v1,a1/3);

	Ve v2=B-C;

	db a2=Angle(A-C,v2);

	v2=Rotate(v2,-a2/3);//负数表示顺时针旋转

	return GetIntersection(B,v1,C,v2);

}

void print(Po P){printf("%.6lf %.6lf ",P.x,P.y);}

int main(){

	int T;

	scanf("%d",&T);

	Po A,B,C,D,E,F;

	while(T--){

		A=read_p();

		B=read_p();

		C=read_p();

		D=getD(A,B,C);

		E=getD(B,C,A);

		F=getD(C,A,B);

		print(D);print(E);print(F);puts("");

	}

	return 0;

}