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Morley's Theorem UVA - 11178

2017-07-24 15:53 302 查看
向量函数板

题意:给你一个三角形的三个点abc,求他们的三等分点的交点

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <vector>
using namespace std;
const int N = 13;
const int M= 1<<N;
int st[M],ma[M];
int dp
[M];

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

typedef point Vector;

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

double dot(Vector a,Vector b)//点乘
{
return a.x*b.x+a.y*b.y;
}

double cross(Vector a,Vector b)//叉乘
{
return a.x*b.y-a.y*b.x;
}

double len (Vector a)
{
return sqrt(a.x*a.x+a.y*a.y);
}

Vector rotate1(Vector a,double rad)//逆时针旋转rad
{
return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}

point getans (point p,Vector v,point q,Vector w)//求两条直线的交点,唯一交点,当且仅当不共线
{
Vector u=p-q;
double t=cross(w,u)/cross(v,w);
return p+v*t;
}

point getPoint(point a,point b,point c)
{
Vector bc=c-b;
Vector ba=a-b;
double rad=acos(dot(bc,ba)/len(bc)/len(ba));
Vector bd=rotate1(bc,rad/3);

Vector cb=b-c;
Vector ca=a-c;
double rad1=acos(dot(cb,ca)/len(cb)/len(ca));
Vector cd=rotate1(cb,-rad1/3);

return getans(b,bd,c,cd);

}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
point a,b,c;
scanf("%lf %lf %lf %lf %lf %lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
point d=getPoint(a,b,c);//后两个的 顺序不能写反了,因为有顺时针逆时针的方向问题
point e=getPoint(b,c,a);
point f=getPoint(c,a,b);
printf("%.6f %.6f ",d.x,d.y);
printf("%.6f %.6f ",e.x,e.y);
printf("%.6f %.6f\n",f.x,f.y);
}
return 0;
}
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