題目大意:求一個三角形中每個內角的角三等分線組成的三角形的三個點的座標;
題目解析:沒有算法可言,直接上模板;
AC代碼:
#include<bits/stdc++.h>
using namespace std;
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) {return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A,Vector 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);
}
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;} //點的點積
double Length(Vector A) {return sqrt(Dot(A,A));} //向量的長度
double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));} //向量之間的角度
double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;} //點的叉積
double Area2(Point A,Point B,Point C){return Cross(B-A,C-A);} //三點構成的三角形面積的兩倍
Vector Rotate(Vector A,double rad) {return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));} //向量逆時針旋轉
Vector Normal(Vector A) //向量的法線
{
double L = Length(A);
return Vector(-A.y/L,A.x/L);
}
//定義直線P+tv,計算兩直線的交點,前提是兩直線不平行
Point GetLineIntersection(Point P,Point v,Point Q,Point w)
{
Vector u=P-Q;
double t=Cross(w,u)/Cross(v,w);
return P+v*t;
}
//點到直線的距離
double DistanceToLine(Point P,Point A,Point B)
{
Vector v1=B-A,v2=P-A;
return fabs(Cross(v1,v2))/Length(v1);
}
//點到線段的距離
double DistanceToSegement(Point P,Point A,Point B)
{
if(A==B) return Length(P-A);
Vector v1=B-A,v2=P-A,v3=P-B;
if(dcmp(Dot(v1,v2))<0) return Length(v2);
else if(dcmp(Dot(v1,v3))>0) return Length(v3);
else return fabs(Cross(v1,v2))/Length(v1);
}
//點在直線上的投影
Point GetLineProjection(Point P,Point A,Point B)
{
Vector v=B-A;
return A+v*(Dot(v,P-A)/Dot(v,v));
}
//判斷兩直線是否規範相交
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
//計算多邊形的有向面積
double PolygonArea(Point* p,int n)
{
double area=0;
for(int i=1;i<n-1;i++)
{
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
}
return area/2;
}
Point fun(Point a,Point b,Point c)
{
double ang=Angle(b-a,c-a);
Point v1=Rotate(b-a,ang/3);
ang=Angle(a-b,c-b);
Point v2=Rotate(a-b,-ang/3);
return GetLineIntersection(a,v1,b,v2);
}
int main()
{
Point a,b,c,d,e,f;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf",&a.x,&a.y);
scanf("%lf%lf",&b.x,&b.y);
scanf("%lf%lf",&c.x,&c.y);
d=fun(b,c,a);
e=fun(c,a,b);
f=fun(a,b,c);
printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",d.x,d.y,e.x,e.y,f.x,f.y);
}
return 0;
}