按照刘汝佳老师说的,这道题本身没有什么算法可言,
主要是考察选手对于几何算法的应用,
我们已经知道了点A,B,C
如果要求点D的话
我们可以先求出向量C-B的坐标,然后求出向量C-B与A-B的夹角。
再把夹角/3,这样我们就找到了∠CBD的度数。
再把向量C-B逆时针旋转∠CBD度
就求出了点D的坐标,
E,F同理
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<cmath>
5 #define Vector Point
6 using namespace std;
7 inline void read(int &n)
8 {
9 char c='+';bool flag=0;n=0;
10 while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
11 while(c>='0'&&c<='9') n=n*10+(c-48),c=getchar();
12 if(flag==1)n=-n;
13 }
14 const double PI=acos(-1);
15 const double eps=1e-10;
16 int dcmp(double x) {return (fabs(x)<eps)?0:(x<0?-1:1);}
17 struct Point
18 {
19 double x,y;
20 Point(double x=0,double y=0):x(x),y(y){};
21 };
22 Vector operator + (Vector A,Vector B) {return Vector(A.x + B.x,A.y + B.y);}
23 Vector operator - (Vector A,Vector B) {return Vector(A.x - B.x,A.y - B.y);}
24 Vector operator * (Vector A,double P) {return Vector(A.x * P,A.y * P);}
25 Vector operator / (Vector A,double P) {return Vector(A.x / P,A.y / P);}
26 bool operator < (const Point &a,const Point &b){return a.x < b.x || (a.x == b.x && a.y < b.y);}
27 bool operator == (const Point &a,const Point &b){return dcmp(a.x - b.x)==0 && dcmp(a.y - b.y)==0;}
28
29 double Dot(Vector A,Vector B){return A.x * B.x + A.y * B.y;}//点积 ********************
30 double Length(Vector A){return sqrt(Dot(A,A));}// 求向量的长度
31 double Angle(Vector A,Vector B){return acos(Dot(A,B) / (Length(A)) / Length(B));} //求两个向量的夹角
32
33 double Cross(Vector A,Vector B){return A.x * B.y-A.y * B.x;}// 两个向量的叉积
34 double Area2(Point A,Point B,Point C){return Cross(B - A,C - A);}
35
36 Vector Rotate(Vector A,double rad){return Vector(A.x * cos(rad)-A.y * sin(rad),A.x * sin(rad)+A.y * cos(rad));}// 向量旋转*********
37
38 Point GetLineIntersection(Point P,Point v,Point Q,Point w)// 两直线的交点 ****************
39 {
40 Vector u=P-Q;
41 double t=Cross(w,u)/Cross(v,w);
42 return P+v*t;
43 }
44
45 double DistanceToLine(Point P,Point A,Point B)// 点P到直线AB的距离
46 {
47 Vector v1=B - A;Vector v2= P-A;
48 return fabs(Cross(v1,v2)) / Length(v1);
49 }
50
51 double DistanceToSegment(Point P,Point A,Point B)// 点P到线段AB的距离
52 {
53 if(A==B) return Length(P-A);
54 Vector v1=B-A, v2=P-A, v3=P-B;
55 if(dcmp(Dot(v1,v2)) < 0) return Length(v2);
56 else if(dcmp(Dot(v1,v3)) > 0) return Length(v3);// 怎么会是大于。。。。。。
57 else return fabs(Cross(v1,v2)) / Length(v1);
58 }
59
60 Point GetLineProjection(Point P, Point A, Point B)// 求点P在直线AB上的正投影
61 {
62 Vector v=B-A;
63 return A+v*(Dot(v, P-A) / Dot(v,v));
64 }
65
66 bool SegmentProperIntersection(Point a1, Point a2, Point b1,Point b2)//判断两直线是否相交
67 {
68 double c1 = Cross(a2-a1,b1-a1) , c2 = Cross(a2-a1,b2-a1),
69 c3 = Cross(b2-b1,a1-b1) , c4 = Cross(b2-b1,a2-b1);
70 return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
71 }
72
73 double PolygonArae(Point *p,int n)// 求多边形的有向面积
74 {
75 double area = 0;
76 for(int i=1;i<=n-1;i++)
77 area += Cross(p[i]-p[0],p[i+1]-p[0]);
78 return area/2;
79 }
80 Point read_point()
81 {
82 double x,y;
83 scanf("%lf%lf",&x,&y);
84 return Point(x,y);
85 }
86 Point getans(Point A,Point B,Point C)
87 {
88 Vector v1= C-B;
89 double ang1=Angle(A-B,v1);
90 v1=Rotate(v1,ang1/3);
91
92 Vector v2= B-C;
93 double ang2=Angle(A-C,v2);
94 v2=Rotate(v2,-ang2/3);
95
96 return GetLineIntersection(B,v1,C,v2);
97 }
98 int main()
99 {
100 int T;read(T);
101 while(T--)
102 {
103 Point A,B,C,D,E,F;
104 A=read_point();
105 B=read_point();
106 C=read_point();
107 D=getans(A,B,C);
108 E=getans(B,C,A);
109 F=getans(C,A,B);
110 printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n", D.x, D.y, E.x, E.y, F.x, F.y);
111 }
112
113 return 0;
114 }