给定一组固定的点。我们需要找到给定集合的凸包。当一个点从集合中移除时，我们还需要找到凸包。
例子：
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)
先决条件：凸包（简单的分而治之算法）
解决上述问题的算法非常简单。我们只是检查要删除的点是否是凸包的一部分。如果是，那么我们必须从初始集合中删除该点，然后再次制作凸包（请参阅凸包（分而治之））。
如果没有，那么我们已经有了解决方案（凸包不会改变）。
// C++ program to demonstrate delete operation
// on Convex Hull.
#include
using namespace std;
// stores the center of polygon (It is made
// global becuase it is used in comare function)
pair mid;
// determines the quadrant of a point
// (used in compare())
int quad(pair 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 a, pair b,
pair 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 p1, pair q1)
{
pair p = make_pair(p1.first - mid.first,
p1.second - mid.second);
pair 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 > merger(vector > a,
vector > 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 a[ia].first)
ia = i;
// ib -> leftmost point of b
for (int i=1; i=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 > 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 > bruteHull(vector > 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 >s;
for (int i=0; i= 0)
pos++;
}
if (pos == a.size() || neg == a.size())
{
s.insert(a[i]);
s.insert(a[j]);
}
}
}
vector >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> findHull(vector> 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>left, right;
for (int i=0; i>left_hull = findHull(left);
vector>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> removePoint(vector> a,
vector> hull,
pair 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;
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 >hull = findHull(a);
cout << "Convex hull:\n";
for (auto e : hull)
cout << e.first << " "
<< e.second << endl;
pair 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;
}
输出：
convex hull:
-3 0
-1 -9
2 -6
5 3
2 5
时间复杂度：
很容易看出，每个查询所花费的最长时间是构造凸包所花费的时间，为 O(n*logn)。因此，整体复杂度为 O(q*n*logn)，其中 q 是要删除的点数。
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