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Convex Hull Graham Scan in C++
In this tutorial, we will be discussing a program to find the convex hull of a given set of points.
Convex hull is the smallest polygon convex figure containing all the given points either on the boundary on inside the figure.
In Graham Scan, firstly the pointes are sorted to get to the bottommost point. Then the points are traversed in order and discarded or accepted to be on the boundary on the basis of their order.
Example
#include <iostream>
#include <stack>
#include <stdlib.h>
using namespace std;
struct Point{
int x, y;
};
//point reference for sorting other points
Point p0;
//moving to the next top in stack
Point nextToTop(stack<Point> &S){
Point p = S.top();
S.pop();
Point res = S.top();
S.push(p);
return res;
}
//swapping two points
int swap(Point &p1, Point &p2){
Point temp = p1;
p1 = p2;
p2 = temp;
}
//calculating the square of difference
int distSq(Point p1, Point p2){
return (p1.x - p2.x)*(p1.x - p2.x) +
(p1.y - p2.y)*(p1.y - p2.y);
}
//checking the orientation of points
int orientation(Point p, Point q, Point r){
int val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if (val == 0) return 0;
return (val > 0)? 1: 2;
}
//sorting and comparing the points
int compare(const void *vp1, const void *vp2){
Point *p1 = (Point *)vp1;
Point *p2 = (Point *)vp2;
int o = orientation(p0, *p1, *p2);
if (o == 0)
return (distSq(p0, *p2) >= distSq(p0, *p1))? -1 : 1;
return (o == 2)? -1: 1;
}
//printing convex hull
void convexHull(Point points[], int n){
int ymin = points[0].y, min = 0;
for (int i = 1; i < n; i++){
int y = points[i].y;
if ((y < ymin) || (ymin == y &&
points[i].x < points[min].x))
ymin = points[i].y, min = i;
}
swap(points[0], points[min]);
p0 = points[0];
qsort(&points[1], n-1, sizeof(Point), compare);
for (int i=1; i<n; i++){
while (i < n-1 && orientation(p0, points[i],
points[i+1]) == 0)
i++;
points[m] = points[i];
m++; //updating size of modified array
}
if (m < 3) return;
stack<Point> S;
S.push(points[0]);
S.push(points[1]);
S.push(points[2]);
for (int i = 3; i < m; i++){
while (orientation(nextToTop(S), S.top(), points[i]) != 2)
S.pop();
S.push(points[i]);
}
while (!S.empty()){
Point p = S.top();
cout << "(" << p.x << ", " << p.y <<")" << endl;
S.pop();
}
}
int main(){
Point points[] = {{0, 3}, {1, 1}, {2, 2}, {4, 4},
{0, 0}, {1, 2}, {3, 1}, {3, 3}};
int n = sizeof(points)/sizeof(points[0]);
convexHull(points, n);
return 0;
}Output
(0, 3) (4, 4) (3, 1) (0, 0)
Updated on: 29-Jan-2020
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