Perimeter of Convex hull for a given set of points

Given n 2-D points points[], the task is to find the perimeter of the convex hull for the set of points. A convex hull for a set of points is the smallest convex polygon that contains all the points.

Examples:

Input: points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1}, {3, 0}, {0, 0}, {3, 3}}
Output: 12



Input: points[] = {{0, 2}, {2, 1}, {3, 1}, {3, 7}}
Output: 15.067

Approach: Monotone chain algorithm constructs the convex hull in O(n * log(n)) time. We have to sort the points first and then calculate the upper and lower hulls in O(n) time. The points will be sorted with respect to x-coordinates (with respect to y-coordinates in case of a tie in x-coordinates), we will then find the left most point and then try to rotate in clockwise direction and find the next point and then repeat the step until we reach the rightmost point and then again rotate in the the clockwise direction and find the lower hull.
We will then find the perimeter of the convex hull using the points on the convex hull which can be done in O(n) time as the points are already sorted in clockwise order.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
#define llu long long int
using namespace std;
  
struct Point {
  
    llu x, y;
  
    bool operator<(Point p)
    {
        return x < p.x || (x == p.x && y < p.y);
    }
};
  
// Cross product of two vectors OA and OB
// returns positive for counter clockwise
// turn and negative for clockwise turn
llu cross_product(Point O, Point A, Point B)
{
    return (A.x - O.x) * (B.y - O.y)
           - (A.y - O.y) * (B.x - O.x);
}
  
// Returns a list of points on the convex hull
// in counter-clockwise order
vector<Point> convex_hull(vector<Point> A)
{
    int n = A.size(), k = 0;
  
    if (n <= 3)
        return A;
  
    vector<Point> ans(2 * n);
  
    // Sort points lexicographically
    sort(A.begin(), A.end());
  
    // Build lower hull
    for (int i = 0; i < n; ++i) {
  
        // If the point at K-1 position is not a part
        // of hull as vector from ans[k-2] to ans[k-1]
        // and ans[k-2] to A[i] has a clockwise turn
        while (k >= 2
               && cross_product(ans[k - 2], 
                            ans[k - 1], A[i]) <= 0)
            k--;
        ans[k++] = A[i];
    }
  
    // Build upper hull
    for (size_t i = n - 1, t = k + 1; i > 0; --i) {
  
        // If the point at K-1 position is not a part
        // of hull as vector from ans[k-2] to ans[k-1]
        // and ans[k-2] to A[i] has a clockwise turn
        while (k >= t
               && cross_product(ans[k - 2], 
                          ans[k - 1], A[i - 1]) <= 0)
            k--;
        ans[k++] = A[i - 1];
    }
  
    // Resize the array to desired size
    ans.resize(k - 1);
  
    return ans;
}
  
// Function to return the distance between two points
double dist(Point a, Point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x)
                + (a.y - b.y) * (a.y - b.y));
}
  
// Function to return the perimeter of the convex hull
double perimeter(vector<Point> ans)
{
    double perimeter = 0.0;
  
    // Find the distance between adjacent points
    for (int i = 0; i < ans.size() - 1; i++) {
        perimeter += dist(ans[i], ans[i + 1]);
    }
  
    // Add the distance between first and last point
    perimeter += dist(ans[0], ans[ans.size() - 1]);
  
    return perimeter;
}
  
// Driver code
int main()
{
    vector<Point> points;
  
    // Add points
    points.push_back({ 0, 3 });
    points.push_back({ 2, 2 });
    points.push_back({ 1, 1 });
    points.push_back({ 2, 1 });
    points.push_back({ 3, 0 });
    points.push_back({ 0, 0 });
    points.push_back({ 3, 3 });
  
    // Find the convex hull
    vector<Point> ans = convex_hull(points);
  
    // Find the perimeter of convex polygon
    cout << perimeter(ans);
  
    return 0;
}

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Output:


12





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