HDU 2857 Mirror and Light [计算几何]

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Description

给一条直线和两个点，这两个点是由那条直线反射得到，求反射的那个点

Algorithm

求对称点
求两直线交点

Code

#include <cstdio>#include <cmath>#include <iostream>using namespace std;const double eps = 1e-10;double add (double a, double b){  if (abs(a + b) < eps * (abs(a) + abs(b))) return 0;  return a + b;}struct P{  double x, y;  P(){}  P(double x, double y):x(x),y(y){}  P operator + (P p)  {    return P(add(x, p.x), add(y, p.y));  }  P operator - (P p)  {    return P(add(x, -p.x), add(y, -p.y));  }  P operator * (double d)  {    return P(x * d, y * d);  }};double det(P p1, P p2){  return add(p1.x * p2.y, -p1.y * p2.x);}bool g_equal(double a, double b){  if (a > b - eps && a < b + eps) return true;  return false;}//求p关于p1, p2所成直线的对称点P symmetric_point(P p, P p1, P p2){  //直线与x轴垂直    if (g_equal(p1.x, p2.x)) return P(2 * p1.x - p.x, p.y);  double k = (p1.y - p2.y ) / (p1.x - p2.x);  if (g_equal(k, 0)) return P(p.x, 2 * p1.y - p.y);  double x = (2*k*k*p1.x + 2*k*p.y - 2*k*p1.y - k*k*p.x + p.x) / (1 + k * k);  double y = p.y - (x - p.x) / k;  return P(x, y);}P intersection(P p1, P p2, P q1, P q2){  return p1 + (p2 - p1) * (det(q2 - q1, q1 - p1) / det(q2 - q1, p2 - p1));}void solve(){  P p1, p2, ps, pe;  cin >> p1.x >> p1.y >> p2.x >> p2.y >> ps.x >> ps.y >> pe.x >> pe.y;  ps = symmetric_point(ps, p1, p2);  P ans = intersection(p1, p2, ps, pe);  printf("%.3lf %.3lf\n", ans.x, ans.y);}int main(){  int t;  cin >> t;  for (int i = 0; i < t; i++)    solve();}

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