UVA 11178

问题描述：

给定三角形的三个点，求三角形角三平分线连成的内部小三角形的三个点的坐标

 

 

这里主要还是几何问题上的大量函数求解问题，

这题的关键是经过两点的确定直线的交点求解问题，我们总是可以利用叉积解决大量问题

通过叉积的比来得到那个方向的向量比例，最后得到点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//用来保证double的精度问题
#define eps 1e-10
const double PI = acos(-1);
struct Point {
    double x, y;
    Point(double x=0,double y=0):x(x),y(y){}//构造函数，方便代码编写
};
typedef Point Vector;

bool operator==(const Point &a,const Point &b)
{
    return abs(a.x-b.x)<eps && abs(a.y-b.y)<eps;
}

Vector operator-(Point a,Point b)
{
    return Vector(a.x-b.x,a.y-b.y);
}

Vector operator+(Vector a,Vector b)
{
    return Vector(b.x+a.x,b.y+a.y);
}

Vector operator*(Vector a,double b)
{
    return Vector(a.x*b,a.y*b);
}

Vector operator/(Vector a,double b)
{
    return Vector(a.x/b,a.y/b);
}


//两向量的内积
double Dot(Vector a,Vector b)
{
    return a.x*b.x + a.y*b.y;
}

//两向量的叉积
double Cross(Vector a,Vector b)
{
    return a.x*b.y - a.y*b.x;
}

//向量的长度
double length(Vector a)
{
    return sqrt(Dot(a,a));
}

//两向量的夹角弧度
double get_rad(Vector a,Vector b)
{
    return acos((Dot(a,b))/length(a)/length(b));
}

//求一个向量逆时针偏转一定弧度后得到的向量，若是顺时针，我们在弧度上加一个负号就行了
Vector get_revovle(Vector a,double rad)
{
    double x = a.x*cos(rad) - a.y*sin(rad);
    double y = a.x*sin(rad) + a.y*cos(rad);
    return Vector(x,y);
}

//得到经过两个点的确定直线的交点
Point getLineIntersection(Point p,Vector vp,Point q,Vector vq)
{
    Vector u = p-q;
    double t = Cross(vq,u) / Cross(vp,vq);
    return p+vp*t;
}

//这个函数求以a为顶点，b，c延伸出1/3弧度处的直线的交点
Point getIntersection(Point a,Point b,Point c)
{
    Vector v1 = a-b;
    Vector v2 = a-c;
    Vector v3 = c-b;
    Vector v4 = b-c;
    double rad1 = get_rad(v1,v3);
    double rad2 = get_rad(v2,v4);
    rad1/=3;
    rad2/=3;
    Vector rev1 = get_revovle(v3,rad1);
    Vector rev2 = get_revovle(v4,-rad2);//负数表示顺时针旋转

    return getLineIntersection(b,rev1,c,rev2);
}

int main()
{
    int T;
    double x1,y1,x2,y2,x3,y3;
    scanf("%d",&T);
    while(T--){
        scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);

        Point A(x1,y1);
        Point B(x2,y2);
        Point C(x3,y3);

        Point ans1 = getIntersection(A,B,C);
        Point ans2 = getIntersection(B,C,A);
        Point ans3 = getIntersection(C,A,B);

        printf("%.6f %.6f %.6f %.6f %.6f %.6f
",ans1.x,ans1.y,ans2.x,ans2.y,ans3.x,ans3.y);
    }
    return 0;
}