题意

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分析

就按题意模拟即可，注意到对称性，只需要知道如何求其中一个。

注意A、B、C按逆时针排列，利用这个性质可以避免旋转时分类讨论。

时间复杂度\(O(T)\)

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
    rg T data=0;
    rg int w=1;
    rg char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data*w;
}
template<class T>T read(T&x)
{
    return x=read<T>();
}
using namespace std;
typedef long long ll;

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

struct Point
{
    double x,y;
    Point(double x=0,double y=0)
    :x(x),y(y){}

    bool operator<(co Point&rhs)co
    {
        return x<rhs.x||(x==rhs.x&&y<rhs.y);
    }

    bool operator==(co Point&rhs)co
    {
        return dcmp(x-rhs.x)==0&&dcmp(y-rhs.y)==0;
    }
};
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);
}

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));
}

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

double Area2(Point A,Point B,Point C)
{
    return Cross(B-A,C-A);
}

Vector Rotate(Vector A,double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}

Vector Normal(Vector A)
{
    double L=Length(A);
    return Vector(-A.y/L,A.x/L);
}

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;
}

double DistanceToLine(Point P,Point A,Point B)
{
    Vector v1=B-A,v2=P-A;
    return fabs(Cross(v1,v2))/Length(v1);
}

double DistanceToSegment(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);
    if(dcmp(Dot(v1,v3))>0)
        return Length(v3);
    return DistanceToLine(P,A,B);
}

Point GetLineProjection(Point P,Point A,Point B)
{
    Vector v=B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}

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;
}

bool OnSegment(Point p,Point a1,Point a2)
{
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}

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 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);

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

//Point read()
//{
//  Point p;
//  scanf("%lf %lf",&p.x,&p.y);
//  return p;
//}

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    int T;
    scanf("%d",&T);
    Point A,B,C,D,E,F;
    while(T--)
    {
//      read(A);read(B);read(C);
        scanf("%lf %lf %lf %lf %lf %lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y);
//      cerr<<"A="<<A.x<<" "<<A.y<<endl;
//      cerr<<"B="<<B.x<<" "<<B.y<<endl;
//      cerr<<"C="<<C.x<<" "<<C.y<<endl;
        D=getD(A,B,C);
        E=getD(B,C,A);
        F=getD(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;
}

Hint

注意向量的读入。开始想重载一个read结果是错的，迫不得已改成直接scanf。

以后可以在结构体里面实现一个read。

UVA11178 Morley's Theorem

原文地址：https://www.cnblogs.com/autoint/p/10125282.html