Dado un conjunto fijo de puntos. Necesitamos encontrar el casco convexo del conjunto dado. También necesitamos encontrar un casco convexo cuando se elimina un punto del conjunto.
Ejemplo:
Initial Set of Points: (-2, 8) (-1, 2) (0, 1) (1, 0) (-3, 0) (-1, -9) (2, -6) (3, 0) (5, 3) (2, 5) Initial convex hull:- (-2, 8) (-3, 0) (-1, -9) (2, -6) (5, 3) Point to remove from the set : (-2, 8) Final convex hull: (2, 5) (-3, 0) (-1, -9) (2, -6) (5, 3)
Requisito previo: casco convexo (algoritmo simple divide y vencerás)
El algoritmo para resolver el problema anterior es muy fácil. Simplemente verificamos si el punto a eliminar es parte del casco convexo. Si es así, entonces tenemos que eliminar ese punto del conjunto inicial y luego hacer el casco convexo nuevamente (consulte Casco convexo (divide y vencerás) ).
Y si no, ya tenemos la solución (el casco convexo no cambiará).
C++
// C++ program to demonstrate delete operation // on Convex Hull. #include<bits/stdc++.h> using namespace std; // stores the center 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>> findHull(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 = findHull(left); vector<pair<int, int>>right_hull = findHull(right); // merging the convex hulls return merger(left_hull, right_hull); } // Returns the convex hull for the given set of points after // remviubg a point p. vector<pair<int, int>> removePoint(vector<pair<int, int>> a, vector<pair<int, int>> hull, pair<int, int> p) { // checking whether the point is a part of the // convex hull or not. bool found = 0; for (int i=0; i < hull.size() && !found; i++) if (hull[i].first == p.first && hull[i].second == p.second) found = 1; // If point is not part of convex hull if (found == 0) return hull; // if it is the part of the convex hull then // we remove the point and again make the convex hull // and if not, we print the same convex hull. for (int i=0; i<a.size(); i++) { if (a[i].first==p.first && a[i].second==p.second) { a.erase(a.begin()+i); break; } } sort(a.begin(), a.end()); return findHull(a); } // 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> >hull = findHull(a); cout << "Convex hull:\n"; for (auto e : hull) cout << e.first << " " << e.second << endl; pair<int, int> p = make_pair(-5, -3); removePoint(a, hull, p); cout << "\nModified Convex Hull:\n"; for (auto e:hull) cout << e.first << " " << e.second << endl; return 0; }
Producción:
convex hull: -3 0 -1 -9 2 -6 5 3 2 5
Complejidad del tiempo:
es simple ver que el tiempo máximo que se tarda por consulta es el tiempo que se tarda en construir la envolvente convexa que es O(n*logn). Entonces, la complejidad general es O(q*n*logn), donde q es el número de puntos a eliminar.
Espacio Auxiliar: O(n), ya que se han tomado n espacios extra.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA