Uva 11178 Morley's Theorem 向量旋转+求直线交点

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=9

题意：

Morlery定理是这样的：作三角形ABC每个内角的三等分线。相交成三角形DEF。则DEF为等边三角形，你的任务是给你A,B,C点坐标求D,E,F的坐标

思路:

根据对称性，我们只要求出一个点其他点一样：我们知道三点的左边即可求出每个夹角，假设求Ｄ，我们只要将向量BC

旋转rad/3的到直线BD，然后旋转向量CB然后得到CD，然后就是求两直线的交点了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val) memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("d.out", "w", stdout)
#define ll unsigned long long
#define keyTree (chd[chd[root][1]][0])

#define M 100007
#define N 300017

using namespace std;

const double eps = 1e-;
const int inf = 0x7f7f7f7f;
const int mod = ;

struct Point
{
    double x,y;
    Point(double tx = ,double ty = ) : x(tx),y(ty){}
};
typedef Point Vtor;
//向量的加减乘除
Vtor operator + (Vtor A,Vtor B) { return Vtor(A.x + B.x,A.y + B.y); }
Vtor operator - (Point A,Point B) { return Vtor(A.x - B.x,A.y - B.y); }
Vtor operator * (Vtor A,double p) { return Vtor(A.x*p,A.y*p); }
Vtor operator / (Vtor A,double p) { return Vtor(A.x/p,A.y/p); }
bool operator < (Point A,Point B) { return A.x < B.x || (A.x == B.x && A.y < B.y);}
int dcmp(double x){ if (fabs(x) < eps) return ; else return x <  ? - : ; }
bool operator == (Point A,Point B) {return dcmp(A.x - B.x) ==  && dcmp(A.y - B.y) == ; }
//向量的点积，长度，夹角
double Dot(Vtor A,Vtor B) { return A.x*B.x + A.y*B.y; }
double Length(Vtor A) { return sqrt(Dot(A,A)); }
double Angle(Vtor A,Vtor B) { return acos(Dot(A,B)/Length(A)/Length(B)); }
//叉积，三角形面积
double Cross(Vtor A,Vtor B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A,Point B,Point C) { return Cross(A - B,C - B); }
//向量的旋转,求向量的单位法线（即左转90度，然后长度归一）
Vtor Rotate(Vtor A,double rad){ return Vtor(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad)); }
Vtor Normal(Vtor A)
{
    double L = Length(A);
    return Vtor(-A.y/L, A.x/L);
}
//直线的交点
Point GetLineIntersection(Point P,Vtor v,Point Q,Vtor w)
{
    Vtor u = P - Q;
    double t = Cross(w,u)/Cross(v,w);
    return P + v*t;
}
//点到直线的距离
double DistanceToLine(Point P,Point A,Point B)
{
    Vtor v1 = B - A;
    return Cross(P,v1)/Length(v1);
}
//点到线段的距离
double DistanceToSegment(Point P,Point A,Point B)
{
    if (A == B) return Length(P - A);
    Vtor v1 =  B - A , v2 = P - A, v3 = P - B;
    if (dcmp(Dot(v1,v2)) < ) return Length(P - A);
    else if (dcmp(Dot(v1,v3)) > ) return Length(P - B);
    else return Cross(v1,v2)/Length(v1);
}
//点到直线的映射
Point GetLineProjection(Point P,Point A,Point B)
{
    Vtor v = B - A;
    return A + v*Dot(v,P - A)/Dot(v,v);
}

//判断线段是否规范相交
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(a1 - a2,b1 - a1), c4 = Cross(a1 - a2,b2 - a2);
    return dcmp(c1)*dcmp(c2) <  && dcmp(c3)*dcmp(c4) < ;
}
//判断点是否在一条线段上
bool OnSegment(Point P,Point a1,Point a2)
{
    return dcmp(Cross(a1 - P,a2 - P)) ==  && dcmp(Dot(a1 - P,a2 - P)) < ;
}
//多边形面积
double PolgonArea(Point *p,int n)
{
    double area = ;
    for (int i = ; i < n - ; ++i)
    area += Cross(p[i] - p[],p[i + ] - p[]);
    return area/;
}

Point solve(Point A,Point B,Point C)
{
    double rad1 = Angle(A - B, C - B)/3.0;
    Vtor v1 = C - B;
    v1 = Rotate(v1,rad1);

    double rad2 = Angle(A - C,B - C)/3.0;
    Vtor v2 = B - C;
    v2 = Rotate(v2,-rad2);

    return GetLineIntersection(B,v1,C,v2);
}
int main()
{

//    Read();
    int T;
    Point A,B,C;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y);
        Point D = solve(A,B,C);
        Point E = solve(B,C,A);
        Point F = solve(C,A,B);

        printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",D.x,D.y,E.x,E.y,F.x,F.y);
    }
    return ;
}