https://www.geeksforgeeks.org/convex-hull-set-1-jarviss-algorithm-or-wrapping/

这个方法很不错，可以看看

 

We strongly recommend to see the following post first.
How to check if two given line segments intersect?

The idea of Jarvis’s Algorithm is simple, we start from the leftmost point (or point with minimum x coordinate value) and we keep wrapping points in counterclockwise direction. The big question is, given a point p as current point, how to find the next point in output? The idea is to use orientation() here. Next point is selected as the point that beats all other points at counterclockwise orientation, i.e., next point is q if for any other point r, we have “orientation(p, r, q) = counterclockwise”. Following is the detailed algorithm.

1) Initialize p as leftmost point.
2) Do following while we don’t come back to the first (or leftmost) point.
…..a) The next point q is the point such that the triplet (p, q, r) is counterclockwise for any other point r.
…..b) next[p] = q (Store q as next of p in the output convex hull).
…..c) p = q (Set p as q for next iteration).

 

 

// A C++ program to find convex hull of a set of points. Refer

// https://www.geeksforgeeks.org/orientation-3-ordered-points/

// for explanation of orientation()

#include <bits/stdc++.h>

using namespace std;

 

struct Point

{

    int x, 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; // clock or counterclock wise

}

 

// Prints convex hull of a set of n points.

void convexHull(Point points[], int n)

{

    // There must be at least 3 points

    if (n < 3) return;

 

    // Initialize Result

    vector<Point> hull;

 

    // Find the leftmost point

    int l = 0;

    for (int i = 1; i < n; i++)

        if (points[i].x < points[l].x)

            l = i;

 

    // Start from leftmost point, keep moving counterclockwise

    // until reach the start point again.  This loop runs O(h)

    // times where h is number of points in result or output.

    int p = l, q;

    do

    {

        // Add current point to result

        hull.push_back(points[p]);

 

        // Search for a point 'q' such that orientation(p, x,

        // q) is counterclockwise for all points 'x'. The idea

        // is to keep track of last visited most counterclock-

        // wise point in q. If any point 'i' is more counterclock-

        // wise than q, then update q.

        q = (p+1)%n;

        for (int i = 0; i < n; i++)

        {

           // If i is more counterclockwise than current q, then

           // update q

           if (orientation(points[p], points[i], points[q]) == 2)

               q = i;

        }

 

        // Now q is the most counterclockwise with respect to p

        // Set p as q for next iteration, so that q is added to

        // result 'hull'

        p = q;

 

    } while (p != l);  // While we don't come to first point

 

    // Print Result

    for (int i = 0; i < hull.size(); i++)

        cout << "(" << hull[i].x << ", "

              << hull[i].y << ")\n";

}

 

// Driver program to test above functions

int main()

{

    Point points[] = {{0, 3}, {2, 2}, {1, 1}, {2, 1},

                      {3, 0}, {0, 0}, {3, 3}};

    int n = sizeof(points)/sizeof(points[0]);

    convexHull(points, n);

    return 0;

}