Casco convexo utilizando el algoritmo Divide and Conquer – Part 1

CPP

// A divide and conquer program to find convex
// hull of a given set of points.
#include<bits/stdc++.h>
using namespace std;
 
// stores the centre of polygon (It is made
// global because it is used in compare function)
pair<int, int> mid;
 
// determines the quadrant of a point
// (used in compare())
int quad(pair<int, int> p)
{
    if (p.first >= 0 && p.second >= 0)
        return 1;
    if (p.first <= 0 && p.second >= 0)
        return 2;
    if (p.first <= 0 && p.second <= 0)
        return 3;
    return 4;
}
 
// Checks whether the line is crossing the polygon
int orientation(pair<int, int> a, pair<int, int> b,
                pair<int, int> c)
{
    int res = (b.second-a.second)*(c.first-b.first) -
              (c.second-b.second)*(b.first-a.first);
 
    if (res == 0)
        return 0;
    if (res > 0)
        return 1;
    return -1;
}
 
// compare function for sorting
bool compare(pair<int, int> p1, pair<int, int> q1)
{
    pair<int, int> p = make_pair(p1.first - mid.first,
                                 p1.second - mid.second);
    pair<int, int> q = make_pair(q1.first - mid.first,
                                 q1.second - mid.second);
 
    int one = quad(p);
    int two = quad(q);
 
    if (one != two)
        return (one < two);
    return (p.second*q.first < q.second*p.first);
}
 
// Finds upper tangent of two polygons 'a' and 'b'
// represented as two vectors.
vector<pair<int, int>> merger(vector<pair<int, int> > a,
                              vector<pair<int, int> > b)
{
    // n1 -> number of points in polygon a
    // n2 -> number of points in polygon b
    int n1 = a.size(), n2 = b.size();
 
    int ia = 0, ib = 0;
    for (int i=1; i<n1; i++)
        if (a[i].first > a[ia].first)
            ia = i;
 
    // ib -> leftmost point of b
    for (int i=1; i<n2; i++)
        if (b[i].first < b[ib].first)
            ib=i;
 
    // finding the upper tangent
    int inda = ia, indb = ib;
    bool done = 0;
    while (!done)
    {
        done = 1;
        while (orientation(b[indb], a[inda], a[(inda+1)%n1]) >=0)
            inda = (inda + 1) % n1;
 
        while (orientation(a[inda], b[indb], b[(n2+indb-1)%n2]) <=0)
        {
            indb = (n2+indb-1)%n2;
            done = 0;
        }
    }
 
    int uppera = inda, upperb = indb;
    inda = ia, indb=ib;
    done = 0;
    int g = 0;
    while (!done)//finding the lower tangent
    {
        done = 1;
        while (orientation(a[inda], b[indb], b[(indb+1)%n2])>=0)
            indb=(indb+1)%n2;
 
        while (orientation(b[indb], a[inda], a[(n1+inda-1)%n1])<=0)
        {
            inda=(n1+inda-1)%n1;
            done=0;
        }
    }
 
    int lowera = inda, lowerb = indb;
    vector<pair<int, int>> ret;
 
    //ret contains the convex hull after merging the two convex hulls
    //with the points sorted in anti-clockwise order
    int ind = uppera;
    ret.push_back(a[uppera]);
    while (ind != lowera)
    {
        ind = (ind+1)%n1;
        ret.push_back(a[ind]);
    }
 
    ind = lowerb;
    ret.push_back(b[lowerb]);
    while (ind != upperb)
    {
        ind = (ind+1)%n2;
        ret.push_back(b[ind]);
    }
    return ret;
 
}
 
// Brute force algorithm to find convex hull for a set
// of less than 6 points
vector<pair<int, int>> bruteHull(vector<pair<int, int>> a)
{
    // Take any pair of points from the set and check
    // whether it is the edge of the convex hull or not.
    // if all the remaining points are on the same side
    // of the line then the line is the edge of convex
    // hull otherwise not
    set<pair<int, int> >s;
 
    for (int i=0; i<a.size(); i++)
    {
        for (int j=i+1; j<a.size(); j++)
        {
            int x1 = a[i].first, x2 = a[j].first;
            int y1 = a[i].second, y2 = a[j].second;
 
            int a1 = y1-y2;
            int b1 = x2-x1;
            int c1 = x1*y2-y1*x2;
            int pos = 0, neg = 0;
            for (int k=0; k<a.size(); k++)
            {
                if (a1*a[k].first+b1*a[k].second+c1 <= 0)
                    neg++;
                if (a1*a[k].first+b1*a[k].second+c1 >= 0)
                    pos++;
            }
            if (pos == a.size() || neg == a.size())
            {
                s.insert(a[i]);
                s.insert(a[j]);
            }
        }
    }
 
    vector<pair<int, int>>ret;
    for (auto e:s)
        ret.push_back(e);
 
    // Sorting the points in the anti-clockwise order
    mid = {0, 0};
    int n = ret.size();
    for (int i=0; i<n; i++)
    {
        mid.first += ret[i].first;
        mid.second += ret[i].second;
        ret[i].first *= n;
        ret[i].second *= n;
    }
    sort(ret.begin(), ret.end(), compare);
    for (int i=0; i<n; i++)
        ret[i] = make_pair(ret[i].first/n, ret[i].second/n);
 
    return ret;
}
 
// Returns the convex hull for the given set of points
vector<pair<int, int>> divide(vector<pair<int, int>> a)
{
    // If the number of points is less than 6 then the
    // function uses the brute algorithm to find the
    // convex hull
    if (a.size() <= 5)
        return bruteHull(a);
 
    // left contains the left half points
    // right contains the right half points
    vector<pair<int, int>>left, right;
    for (int i=0; i<a.size()/2; i++)
        left.push_back(a[i]);
    for (int i=a.size()/2; i<a.size(); i++)
        right.push_back(a[i]);
 
    // convex hull for the left and right sets
    vector<pair<int, int>>left_hull = divide(left);
    vector<pair<int, int>>right_hull = divide(right);
 
    // merging the convex hulls
    return merger(left_hull, right_hull);
}
 
// Driver code
int main()
{
    vector<pair<int, int> > a;
    a.push_back(make_pair(0, 0));
    a.push_back(make_pair(1, -4));
    a.push_back(make_pair(-1, -5));
    a.push_back(make_pair(-5, -3));
    a.push_back(make_pair(-3, -1));
    a.push_back(make_pair(-1, -3));
    a.push_back(make_pair(-2, -2));
    a.push_back(make_pair(-1, -1));
    a.push_back(make_pair(-2, -1));
    a.push_back(make_pair(-1, 1));
 
    int n = a.size();
 
    // sorting the set of points according
    // to the x-coordinate
    sort(a.begin(), a.end());
    vector<pair<int, int> >ans = divide(a);
 
    cout << "convex hull:\n";
    for (auto e:ans)
       cout << e.first << " "
            << e.second << endl;
 
    return 0;
}