Convex Hull

Introduction

The Convex Hull is the subset of points that forms the smallest convex polygon which encloses all points in the set. To visualize this, imagine that each point is a pole. Then, imagine what happens if you were to wrap a rope around the outside of all the poles, and then pull infinitely hard, such that the connections between any two points that lie on the edge of the rope are lines. The set of points that touch the rope is the convex hull.

With Graham Scan

Solution

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#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORE(i, a, b) for (int i = (a); i <= (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define trav(a, x) for (auto &a : x)

#define f first

With Monotone Chain

Solution

With the Monotone Chain algorithm, we start by sorting the given $n$ points in ascending order with respect to their $x$ coordinates. If two points have the same $x$ coordinate, then we will look at the $y$ coordinate.

Next, we will calculate the convex hull in two parts - the upper and the lower hull. Firstly, we observe that the starting and ending points of both upper and lower hulls are the same. They are the points with the lowest and highest $x$ value respectively, $P_0$ and $P_{n-1}$. We start by adding $P_0$ and $P_1$ to the hull. (Note that $P_1$ doesn't necessarily have to be on the convex hull at the end). Then, starting from $P_2$, we iterate through the sorted points and add them to the hull. Let's denote the current point being added as $P_{k}$ and the last point still on the hull as $P_{i}$. When adding new points, we want to make sure that there is no right turn among all segments of the hull, just like in the Graham Scan algorithm discussed above. To achieve this, the segment $P_{i-1}P_i$ should always be on the right side of the segment $P_{i-1}P_k$. This can be calculated by using a cross-product:

• If $(P_i - P_{i-1}) \times (P_k - P_{i-1}) < 0$, the point $P_i$ lies on the left side of the segment $P_{i-1}P_k$. In this case, we have to remove point $P_i$ from the hull and repeat this check.
• If $(P_i - P_{i-1}) \times (P_k - P_{i-1}) = 0$, the point $P_i$ lies on the segment $P_{i-1}P_k$. Whether to include multiple collinear points depends on the question, but for the given question above, we will remove the point $P_i$ as well and repeat the check.
• Otherwise, the point $P_i$ lies on the right side of the segment $P_{i-1}P_k$. In this case, we can add $P_k$ to the hull and process the next point $P_{k+1}$ from the given point list.

After all the points have been processed, we have found the lower hull and will begin to find the upper hull in the same manner. This time, we add point $P_{n-2}$ to the hull and iterate from the end of the points, $P_{n-3}$, to the starting point $P_0$. (The point $P_{n-2}$ also doesn't necessarily have to be the convex hull and could be removed if it causes a right turn).

At the end, we have got all points of the convex hull in the counterclockwise order. To do this in the clockwise order, one only has to change the condition for $(P_i - P_{i-1}) \times (P_k - P_{i-1})$ from more than 0 to less than 0. In this case, the upper hull will be found first and then the lower hull.

This algorithm takes $\mathcal{O}(n \log n)$ time to sort the points and $\mathcal{O}(n)$ time to calculate the hull, giving a final time complexity of $\mathcal{O}(n \log n)$.

An animation of how it works:

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#include <bits/stdc++.h>
using namespace std;

using pii = pair<int, int>;

vector<pii> points;
vector<pii> hull;

// cross product, the signed area of these there points
int area(pii O, pii P, pii Q) {

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import java.io.*;
import java.util.*;

public class ConvexHull {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

		int N = Integer.parseInt(in.readLine());
		while (N > 0) {
			// use a hashset to distinct the points

Rotating Calipers

Solution

Problems