Find Simple Closed Path for a given set of points in C++

Consider we have a set of points. We have to find a simple closed path covering all points. Suppose the points are like below, and the next image is making a closed path on those points.

To get the path, we have to follow these steps −

• Find the bottom left point as P

• Sort other n – 1 point based on the polar angle counterclockwise around P, if polar angle of two points are same, then put them as the distance is shortest

• Traverse the sorted list of points, then make the path

Example

#include <bits/stdc++.h>
using namespace std;
class Point {
   public:
   int x, y;
};
Point p0;
int euclid_dist(Point p1, Point p2) {
   return (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y);
}
int orientation(Point p1, Point p2, Point p3) {
   int val = (p2.y - p1.y) * (p3.x - p2.x) - (p2.x - p1.x) * (p3.y - p2.y);
   if (val == 0) return 0; // colinear
   return (val > 0)? 1: 2; // clockwise or counterclock wise
}
int compare(const void *vp1, const void *vp2) {
   Point *p1 = (Point *)vp1;
   Point *p2 = (Point *)vp2;
   int o = orientation(p0, *p1, *p2);
   if (o == 0)
   return (euclid_dist(p0, *p2) >= euclid_dist(p0, *p1))? -1 : 1;
   return (o == 2)? -1: 1;
}
void findClosedPath(Point points[], int n) {
   int y_min = points[0].y, min = 0;
   for (int i = 1; i < n; i++) {
      int y = points[i].y;
      if ((y < y_min) || (y_min == y && points[i].x < points[min].x))
      y_min = points[i].y, min = i;
   }
   swap(points[0], points[min]);
   p0 = points[0];
   qsort(&points[1], n-1, sizeof(Point), compare); //sort on polar angle
   for (int i=0; i<n; i++)
   cout << "(" << points[i].x << ", "<< points[i].y <<"), ";
}
int main() {
   Point points[] = {{0, 3}, {1, 1}, {2, 2}, {4, 4},{0, 0}, {1, 2}, {3, 1}, {3, 3}};
   int n = sizeof(points)/sizeof(points[0]);
   findClosedPath(points, n);
}

Output

(0, 0), (3, 1), (1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (0, 3),