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 pointif (n < 3){cout << "Convex hull not possible\n";return;}// Finding the point with minimum and// maximum x-coordinateint 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;
}