1 文本格式

// C++ program to implement Quick Hull algorithm
// to find convex hull.
#include<bits/stdc++.h>
using namespace std;

// iPair is integer pairs
#define iPair pair<int, int>

// Stores the result (points of convex hull)
set<iPair> hull;

// Returns the side of point p with respect to line
// joining points p1 and p2.
int findSide(iPair p1, iPair p2, iPair p)
{
    int val = (p.second - p1.second) * (p2.first - p1.first) -
        (p2.second - p1.second) * (p.first - p1.first);

    if (val > 0)
        return 1;
    if (val < 0)
        return -1;
    return 0;
}

// returns a value proportional to the distance
// between the point p and the line joining the
// points p1 and p2
int lineDist(iPair p1, iPair p2, iPair p)
{
    return abs((p.second - p1.second) * (p2.first - p1.first) -
        (p2.second - p1.second) * (p.first - p1.first));
}

// End points of line L are p1 and p2.  side can have value
// 1 or -1 specifying each of the parts made by the line L
void quickHull(iPair a[], int n, iPair p1, iPair p2, int side)
{
    int ind = -1;
    int max_dist = 0;

    // finding the point with maximum distance
    // from L and also on the specified side of L.
    for (int i = 0; i < n; i++)
    {
        int temp = lineDist(p1, p2, a[i]);
        if (findSide(p1, p2, a[i]) == side && temp > max_dist)
        {
            ind = i;
            max_dist = temp;
        }
    }

    // If no point is found, add the end points
    // of L to the convex hull.
    if (ind == -1)
    {
        hull.insert(p1);
        hull.insert(p2);
        return;
    }

    // Recur for the two parts divided by a[ind]
    quickHull(a, n, a[ind], p1, -findSide(a[ind], p1, p2));
    quickHull(a, n, a[ind], p2, -findSide(a[ind], p2, p1));
}

void printHull(iPair a[], int n)
{
    // a[i].second -> y-coordinate of the ith point
    if (n < 3)
    {
        cout << "Convex hull not possible\n";
        return;
    }

    // Finding the point with minimum and
    // maximum x-coordinate
    int min_x = 0, max_x = 0;
    for (int i = 1; i < n; i++)
    {
        if (a[i].first < a[min_x].first)
            min_x = i;
        if (a[i].first > a[max_x].first)
            max_x = i;
    }

    // Recursively find convex hull points on
    // one side of line joining a[min_x] and
    // a[max_x]
    quickHull(a, n, a[min_x], a[max_x], 1);

    // Recursively find convex hull points on
    // other side of line joining a[min_x] and
    // a[max_x]
    quickHull(a, n, a[min_x], a[max_x], -1);

    cout << "The points in Convex Hull are:\n";
    while (!hull.empty())
    {
        cout << "(" << (*hull.begin()).first << ", "
            << (*hull.begin()).second << ") ";
        hull.erase(hull.begin());
    }
}

// Driver code
int main()
{
    iPair a[] = { {0, 3}, {1, 1}, {2, 2}, {4, 4},
               {0, 0}, {1, 2}, {3, 1}, {3, 3} };
    int n = sizeof(a) / sizeof(a[0]);
    printHull(a, n);
    return 0;
}
 

2 代码格式

// C++ program to implement Quick Hull algorithm
// to find convex hull.
#include<bits/stdc++.h>
using namespace std;

// iPair is integer pairs
#define iPair pair<int, int>

// Stores the result (points of convex hull)
set<iPair> hull;

// Returns the side of point p with respect to line
// joining points p1 and p2.
int findSide(iPair p1, iPair p2, iPair p)
{
	int val = (p.second - p1.second) * (p2.first - p1.first) -
		(p2.second - p1.second) * (p.first - p1.first);

	if (val > 0)
		return 1;
	if (val < 0)
		return -1;
	return 0;
}

// returns a value proportional to the distance
// between the point p and the line joining the
// points p1 and p2
int lineDist(iPair p1, iPair p2, iPair p)
{
	return abs((p.second - p1.second) * (p2.first - p1.first) -
		(p2.second - p1.second) * (p.first - p1.first));
}

// End points of line L are p1 and p2.  side can have value
// 1 or -1 specifying each of the parts made by the line L
void quickHull(iPair a[], int n, iPair p1, iPair p2, int side)
{
	int ind = -1;
	int max_dist = 0;

	// finding the point with maximum distance
	// from L and also on the specified side of L.
	for (int i = 0; i < n; i++)
	{
		int temp = lineDist(p1, p2, a[i]);
		if (findSide(p1, p2, a[i]) == side && temp > max_dist)
		{
			ind = i;
			max_dist = temp;
		}
	}

	// If no point is found, add the end points
	// of L to the convex hull.
	if (ind == -1)
	{
		hull.insert(p1);
		hull.insert(p2);
		return;
	}

	// Recur for the two parts divided by a[ind]
	quickHull(a, n, a[ind], p1, -findSide(a[ind], p1, p2));
	quickHull(a, n, a[ind], p2, -findSide(a[ind], p2, p1));
}

void printHull(iPair a[], int n)
{
	// a[i].second -> y-coordinate of the ith point
	if (n < 3)
	{
		cout << "Convex hull not possible\n";
		return;
	}

	// Finding the point with minimum and
	// maximum x-coordinate
	int min_x = 0, max_x = 0;
	for (int i = 1; i < n; i++)
	{
		if (a[i].first < a[min_x].first)
			min_x = i;
		if (a[i].first > a[max_x].first)
			max_x = i;
	}

	// Recursively find convex hull points on
	// one side of line joining a[min_x] and
	// a[max_x]
	quickHull(a, n, a[min_x], a[max_x], 1);

	// Recursively find convex hull points on
	// other side of line joining a[min_x] and
	// a[max_x]
	quickHull(a, n, a[min_x], a[max_x], -1);

	cout << "The points in Convex Hull are:\n";
	while (!hull.empty())
	{
		cout << "(" << (*hull.begin()).first << ", "
			<< (*hull.begin()).second << ") ";
		hull.erase(hull.begin());
	}
}

// Driver code
int main()
{
	iPair a[] = { {0, 3}, {1, 1}, {2, 2}, {4, 4},
			   {0, 0}, {1, 2}, {3, 1}, {3, 3} };
	int n = sizeof(a) / sizeof(a[0]);
	printHull(a, n);
	return 0;
}