uva11178（二维几何计算模板）

题目链接：http://vjudge.net/problem/viewProblem.action?id=18543

#include<iostream>#include<cmath>#include<cstdio>#include<cstring>#include<vector>using namespace std;const double PI = acos(-1.0);struct Point{//点    double x, y;    Point(double _x = 0, double _y = 0):x(_x),y(_y){}};typedef Point Vector;//向量，从代码看就是点的别名//向量+向量=向量，点+向量=点Vector operator + (Vector A, Vector B){    return Vector(A.x+B.x, A.y+B.y);}//点-点=向量Vector operator - (Point A, Point B){    return Vector(A.x-B.x, A.y-B.y);}//向量*数=向量Vector operator * (Vector A, double p){    return Vector(A.x*p, A.y*p);}//向量/数=向量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;}/*基本运算*//*点积,判断夹角大小*/double Dot(Vector A, Vector B){    return A.x*B.x + A.y*B.y;}//长度double Length(Vector A){    return sqrt(Dot(A, A));}//夹角大小a.b = |a||b|cosθdouble Angle(Vector A, Vector B){    return acos(Dot(A, B) / Length(A) / Length(B));}/*叉积  右手定则    1、等于两向量构成四边形的面积    2、判断两向量顺逆时针关系，P*Q>0,Q在P左边        等于0（共线）*/double Cross(Vector A, Vector B){    return A.x*B.y - A.y*B.x;}//平行四边形面积double Area2(Point A, Point B, Point C){    return Cross(B-A, C-A);}//向量旋转Vector Rotate(Vector A, double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));}//直线交点(P点和它的方向,Q点和它的方向)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;}//点到直线的距离，用平行四边形面积除以底double DistanceToLine(Point P, Point A, Point B){    Vector v1 = B - A, v2 = P - A;    return fabs(Cross(v1, v2) / Length(v1));//如果不去绝对值，得有向距离}//点到线段距离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;//向量AB、AP、BP    if(dcmp(Dot(v1, v2)) < 0)return Length(v2);//投影在线段AB左边    else if(dcmp(Dot(v1, v3)) > 0)return Length(v3);//投影在线段AB右边    else return fabs(Cross(v1, v2)) / Length(v1);//P点投影在线段上，距离变为求点到直线距离}//求点在直线上的投影QPoint GetLineProjection(Point P, Point A, Point B){    Vector v = B - A;    return A + v*(Dot(v, P-A) / Dot(v, v));}//线段相交判定bool OnSegment(Point p, Point a1, Point a2){//先判断一个点是否自一条线段上    return dcmp(Cross(a1-p, a2-p))==0 && dcmp(Dot(a1-p, a2-p))<0;}//规范相交，恰有一个交点，每条线段两个端点都在另一个线段的两侧，即叉积符号不同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;}Point read_point(){    Point p;    scanf("%lf%lf",&p.x,&p.y);    return p;}int main(){    int i, j, n, t;    double LA, LB, LC;    Point A, B, C, E, D, F;    Vector BD,CD,CE,AE,AF,BF;    cin>>t;    while(t--)    {        A = read_point();        B = read_point();        C = read_point();//点        LA = Angle(C-A, B-A);        LB = Angle(A-B, C-B);        LC = Angle(B-C, A-C);//角度        BD = Rotate(C-B, LB/3.0);        CD = Rotate(B-C, -LC/3.0);        D = GetLineIntersection(B, BD, C, CD);        CE = Rotate(A-C, LC/3.0);        AE = Rotate(C-A, -LA/3.0);        E = GetLineIntersection(C, CE, A, AE);        AF = Rotate(B-A, LA/3.0);        BF = Rotate(A-B, -LB/3.0);        F = GetLineIntersection(B, BF, A, AF);        printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",               D.x,D.y,E.x,E.y,F.x,F.y);    }    return 0;}