uva 11178

题意：根据A,B,C三点的位置确定D,E,F三个点的位置。

贴模板

  1 #include<cstdio>
  2 #include<cmath>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<iostream>
  6 #include<memory.h>
  7 #include<cstdlib>
  8 #include<vector>
  9 #define clc(a,b) memset(a,b,sizeof(a))
 10 #define LL long long int
 11 using namespace std;
 12 const int N=100010;
 13 const int inf=0x3f3f3f3f;
 14 const double eps = 1e-10;
 15 
 16 struct Point
 17 {
 18     double x, y;
 19     Point(double x = 0, double y = 0) : x(x), y(y) { }
 20 };
 21 
 22 typedef Point Vector;
 23 
 24 Vector operator + (Vector A, Vector B)
 25 {
 26     return Vector(A.x + B.x, A.y + B.y);
 27 }
 28 
 29 Vector operator - (Point A, Point B)
 30 {
 31     return Vector(A.x - B.x, A.y - B.y);
 32 }
 33 
 34 Vector operator * (Vector A, double p)
 35 {
 36     return Vector(A.x * p, A.y * p);
 37 }
 38 
 39 Vector operator / (Vector A, double p)
 40 {
 41     return Vector(A.x / p, A.y / p);
 42 }
 43 
 44 bool operator < (const Point& a, const Point& b)
 45 {
 46     return a.x < b.x || (a.x == b.x && a.y < b.y);
 47 }
 48 
 49 int dcmp(double x)
 50 {
 51     if(fabs(x) < eps) return 0;
 52     return x < 0 ? -1 : 1;
 53 }
 54 
 55 bool operator == (const Point& a, const Point& b)
 56 {
 57     return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
 58 }
 59 
 60 double Dot(Vector A, Vector B)//点乘
 61 {
 62     return A.x * B.x + A.y * B.y;    
 63 }
 64 
 65 double Length(Vector A) //向量的模
 66 {
 67     return sqrt(Dot(A, A));   
 68 }
 69 
 70 double Angle(Vector A, Vector B)//两个向量的夹角
 71 {
 72     return acos(Dot(A, B) / Length(A) / Length(B));    
 73 }
 74 
 75 double Cross(Vector A, Vector B)//叉乘
 76 {
 77     return A.x * B.y - A.y * B.x;    
 78 }
 79 
 80 double Area(Point A, Point B, Point C)//三个点组成的三角形的面积
 81 {
 82     return Cross(B - A, C - A);    
 83 }
 84 
 85 Vector Rotate(Vector A, double rad)   //向量A逆时针旋转rad弧度后的坐标
 86 {
 87     return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
 88 }
 89 
 90 
 91 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)//求直线交点
 92 {
 93     Vector u = P - Q;
 94     double t = Cross(w, u) / Cross(v, w);
 95     return P + v * t;
 96 }
 97 
 98 Point getD(Point A, Point B, Point C)
 99 {
100     Vector v1 = C - B;
101     double a1 = Angle(A-B, v1);
102     v1 = Rotate(v1, a1/3);
103 
104     Vector v2 = B - C;
105     double a2 = Angle(A-C, v2);
106     v2 = Rotate(v2, -a2/3);
107 
108     return GetLineIntersection(B, v1, C, v2);
109 }
110 int main()
111 {
112     int T;
113     Point A, B, C, D, E, F;
114     scanf("%d",&T);
115     while(T--)
116     {
117         scanf("%lf%lf%lf%lf%lf%lf",&A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
118         D = getD(A, B, C);
119         E = getD(B, C, A);
120         F = getD(C, A, B);
121         printf("%lf %lf %lf %lf %lf %lf\n", D.x, D.y, E.x, E.y, F.x, F.y);
122     }
123     return 0;
124 }
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posted @ 2015-10-10 22:10  yyblues  阅读(173)  评论(0)    收藏  举报