例题4.1 Morley定理 UVa11178

1.题目描述：点击打开链接

2.解题思路：本题直接模拟即可。只要知道如何计算D点的坐标，就能算出其他两个点。根据题意，我们需要先计算∠ABC的值a,然后把射线BC逆时针旋转a/3,得到直线BD，同理可以得到直线CD，求交点即可。

3.代码：

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<cassert>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<functional>
using namespace std;

#define me(s)  memset(s,0,sizeof(s))
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair <int, int> P;

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;
}
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));
}

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 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);   //负数表示逆时针旋转
    return GetLineIntersection(B,v1,C,v2);
}

Point read_point()
{
    double x,y;
    scanf("%lf%lf",&x,&y);
    return Point(x,y);
}
int main()
{
    int T;
    Point A,B,C,D,E,F;
    scanf("%d",&T);
    while(T--)
    {
        A=read_point();
        B=read_point();
        C=read_point();
        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);
    }
}

版权声明：本文为博主原创文章，未经博主允许不得转载。