查找给定点集的简单闭合路径

给定一组点，将点连接起来而不交叉。

例子：

Input: points[] = {(0, 3), (1, 1), (2, 2), (4, 4),
                   (0, 0), (1, 2), (3, 1}, {3, 3}};

Output: Connecting points in following order would
        not cause any crossing
       {(0, 0), (3, 1), (1, 1), (2, 2), (3, 3),
        (4, 4), (1, 2), (0, 3)}

强烈建议您最小化浏览器，然后自己尝试。

这个想法是使用排序。

1. 通过比较所有点的y坐标找到最底端的点。如果存在两个具有相同y值的点，则考虑具有较小x坐标值的点。将最低点放在第一个位置。
   
2. 考虑剩余的n-1个点，并按点[0]的逆时针顺序按极角对它们进行排序。如果两点的极角相同，则将最近的点放在第一位。
3. 遍历排序的数组(按角度递增的顺序排序)会产生简单的闭合路径。
   

如何计算角度?
一种解决方案是使用三角函数。
观察：我们不在乎角度的实际值。我们只想按角度排序。
想法：使用方向来比较角度，而无需实际计算角度！

下面是上述想法的C++实现。

// A C++ program to find simple closed path for n points
// for explanation of orientation()
#include 
using namespace std;
  
struct Point
{
    int x, y;
};
  
// A global point needed for  sorting points with reference
// to the first point. Used in compare function of qsort()
Point p0;
  
// A utility function to swap two points
int swap(Point &p1, Point &p2)
{
    Point temp = p1;
    p1 = p2;
    p2 = temp;
}
  
// A utility function to return square of distance between
// p1 and p2
int dist(Point p1, Point p2)
{
    return (p1.x - p2.x)*(p1.x - p2.x) +
           (p1.y - p2.y)*(p1.y - p2.y);
}
  
// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);
  
    if (val == 0) return 0;  // colinear
    return (val > 0)? 1: 2; // clockwise or counterclock wise
}
  
// A function used by library function qsort() to sort
//  an array of points with respect to the first point
int compare(const void *vp1, const void *vp2)
{
   Point *p1 = (Point *)vp1;
   Point *p2 = (Point *)vp2;
  
   // Find orientation
   int o = orientation(p0, *p1, *p2);
   if (o == 0)
     return (dist(p0, *p2) >= dist(p0, *p1))? -1 : 1;
  
   return (o == 2)? -1: 1;
}
  
// Prints simple closed path for a set of n points.
void printClosedPath(Point points[], int n)
{
   // Find the bottommost point
   int ymin = points[0].y, min = 0;
   for (int i = 1; i < n; i++)
   {
     int y = points[i].y;
  
     // Pick the bottom-most. In case of tie, chose the
     // left most point
     if ((y < ymin) || (ymin == y &&
         points[i].x < points[min].x))
        ymin = points[i].y, min = i;
   }
  
   // Place the bottom-most point at first position
   swap(points[0], points[min]);
  
   // Sort n-1 points with respect to the first point.
   // A point p1 comes before p2 in sorted ouput if p2
   // has larger polar angle (in counterclockwise
   // direction) than p1
   p0 = points[0];
   qsort(&points[1], n-1, sizeof(Point), compare);
  
   // Now stack has the output points, print contents
   // of stack
   for (int i=0; i

输出：

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

如果我们使用O(nLogn)排序算法对点进行排序，则上述解决方案的时间复杂度为O(n Log n)。

来源：
http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf