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)