180 lines
4.3 KiB
C++
180 lines
4.3 KiB
C++
#include <cfloat>
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#include <iostream>
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#include <math.h>
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using namespace std;
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struct Vector2d {
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double x, y;
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Vector2d(double a, double b) : x(a), y(b) {}
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Vector2d() : x(.0), y(.0) {}
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Vector2d operator+(Vector2d v) const { return Vector2d(v.x + x, v.y + y); }
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Vector2d operator-(Vector2d v) const { return Vector2d(x - v.x, y - v.y); }
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Vector2d operator-(double a) const { return Vector2d(x - a, y - a); }
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Vector2d operator+(double a) const { return Vector2d(x + a, y + a); }
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Vector2d operator/(double a) const { return Vector2d(x / a, y / a); }
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Vector2d operator*(double a) const { return Vector2d(x * a, y * a); }
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Vector2d operator+=(const Vector2d &v) {
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*this = *this + v;
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return *this;
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}
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double norm() { return sqrt(x * x + y * y); }
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double dot(Vector2d v) { return x * v.x + y * v.y; }
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void reset() {
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x = 0;
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y = 0;
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}
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};
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struct Pedestrian {
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Vector2d curr_pos; // $\vectorarrow{r}_\alpha(t)$
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Vector2d next_pos; // $\vectorarrow{r}_\alpha(t+1)$ position, curr_pos is needed because when calculating the force on pedestrian j, we need the old pos, not the updated
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Vector2d curr_vel, next_vel; // velocity
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Vector2d destination; // $\vectorarrow{r}^0_\alpha$
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Vector2d next_node; // $\vectorarrow{r}^k_\alpha$
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double v0; // speed in m/s e.g. 1.34
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double vMax; // vMax = v0 * 1.3
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// direction Vector2d e = (dest - curr.position).norm() --> calculate on the fly or part of struct?
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void update() { curr_pos = next_pos; }
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};
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// initialize map
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struct Sidewalk{
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int row, col;
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double length, width;
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Sidewalk(int r, int c, double l, double w) : row(r), col(c), length(l), width(w) {}
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/* X = obstacle
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| X |
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| X |
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| X |
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|X |
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*/
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};
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// Numer of pedestrians
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// $n_p$
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const int n_p = 100;
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// Number of iterations
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// $n_{iter}$
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const int n_iter = 100;
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// TODO: Change this
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// $\tau$
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const double tau = 0.12345;
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// F_i is the current position resp. direction vector of pedestrian p in time i?
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void compute_acceleration_term(Vector2d *F_i, Vector2d curr, Vector2d next) {
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// get curr pos --> probably no needed, can be read in F_i
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// get next pos --> call shortest path --> how to encode map?
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// get desired velocity
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// desired speed * desired direction
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// get actual velocity
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// actual speed * actual direction
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// TODO: how to get next node?
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// acceleration * desired speed
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double v_desired = 0.0;
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double v_actual = 0.0;
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double res = (1 / tau) * (v_desired - v_actual);
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}
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Vector2d get_next_node(Pedestrian p, Vector2d map[]) {
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// dummy element
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return Vector2d(p.curr_pos.x + 1.0, p.curr_pos.y + 1.0);
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}
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// return the direction between curr position and destination
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// $\vectorarrow{e}_\alpha(t)$
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Vector2d get_direction(Pedestrian p) {
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Vector2d diff = p.next_node - p.curr_pos;
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double length = diff.norm();
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return (p.next_node - p.curr_pos) / length;
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}
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void compute_repulsive_effects_from_pedestrians(Vector2d *F_i) {
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}
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void compute_repulsive_effects_from_borders(Vector2d *F_i) {
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}
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// Return an array of doubles
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double *w(Vector2d e, Vector2d f) {
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// TODO:
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// $cos(60\degree)$
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// if (e * f > f.norm() * (0.5)) {
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// }
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// return 0;
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}
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int main() {
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// ! DONT TOUCH
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// $\vec{v}^0_i$
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// Velocity at iteration $t$ for pedestrian $i$
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double v[n_iter][n_p];
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// $\vec{r}^0_i$
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// Intended destination at iteration $t$ for pedestrian $i$
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Vector2d r[n_iter][n_p];
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// $\vec{e}_i$
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// Desired direction to the next way point at iteration $t$ for pedestrian $i$
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Vector2d e[n_iter][n_p];
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// $v_i^{max}$
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// Maximum allowed speed for pedestrian $i$
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double v_max[n_p];
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// $v_i^{0}$
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// Desired speed for pedestrian $i$
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double v_0[n_p];
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// $\vec{F}_i$
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// Total social force for pedestrian $i$
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Vector2d F[n_p][n_p];
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// $\vec{r}_0$
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// Desired destination of pedestrian $n_p$
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Vector2d r_0[n_p];
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// for each time frame
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for (int t = 0; t < n_iter; t++) {
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// for each pedestrian
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for (int i = 0; i < n_p; i++) {
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// update...
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for (int itr = 0; itr < n_p; itr++) {
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F[i][itr].reset();
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}
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compute_acceleration_term(F[i]);
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compute_repulsive_effects_from_pedestrians(F[i]);
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compute_repulsive_effects_from_borders(F[i]);
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// TODO: Update $\vec{w}_i$
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// TODO: Update $\vec{v}_i^{t+1}$
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// TODO: Update $\vec{r}_i^{t+1}$
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}
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}
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return 0;
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}
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