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: 12Input: 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:
// C++ implementation of the approach#include <bits/stdc++.h>#define llu long long intusing 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 turnllu 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 ordervector<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 pointsdouble 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 hulldouble 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 codeint 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;} |
12
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.
In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA
