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题意:给定三角形的三个点,让你求它每个角的三等分线所交的顶点。
析:根据自己的以前的数学知识,应该很容易想到思想,比如D点,就是应该求直线BD和CD的交点,
以前还得自己算,现在计算机帮你算,更方便,主要注意的是旋转是顺时针还是逆时针,不要搞错了。
要求BD和CD就得先求那个夹角ABC和ACD,然后三等分。
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = 500 + 10;
const double eps = 1E-10;
struct Point{
double x, y;
Point(double xx = 0, double yy = 0) : x(xx), y(yy) { }
};
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
double Length(Vector A){ return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B){ return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B){ return A.x*B.y - A.y*B.x; }
Vector Rotate(Vector A, double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad)); }
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;
}
Point solve(Point A, Point B, Point C){
double abc = Angle(A-B, C-B) / 3.0;
Vector BD = Rotate(C-B, abc);
double acb = Angle(A-C, B-C) / 3.0;
Vector CD = Rotate(B-C, -acb);
return GetLineIntersection(B, BD, C, CD);
}
int main(){
int T; cin >> T;
Point A, B, C, D, E, F;
double x, y;
while(T--){
scanf("%lf %lf", &x, &y); A = Point(x, y);
scanf("%lf %lf", &x, &y); B = Point(x, y);
scanf("%lf %lf", &x, &y); C = Point(x, y);
D = solve(A, B, C);
E = solve(B, C, A);
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 0;
}
UVa 11178 Morley's Theorem (几何问题)
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5547292.html