在本教程中,我们将讨论一个程序来查找给定点集的凸包。
凸包是最小的多边形凸图,其中包含图内边界上的所有给定点。
在Graham Scan中,首先将点排序到最底点。然后按顺序遍历这些点,并根据它们的顺序将其丢弃或接受到边界上。
#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; }
输出结果
(0, 3) (4, 4) (3, 1) (0, 0)