Jarvis March算法
JarvisMarch算法用于从一组给定的数据点中检测凸包的角点。
从数据集的最左端开始，我们通过逆时针旋转将这些点保留在凸包中。从当前点开始，我们可以通过从当前点检查这些点的方向来选择下一个点。当角度最大时，将选择该点。完成所有点后，当下一个点是起点时，停止算法。
输入输出
Input:
Set of points: {(-7,8), (-4,6), (2,6), (6,4), (8,6), (7,-2), (4,-6), (8,-7),(0,0), (3,- 2),(6,-10),(0,6),(-9,-5),(-8,-2),(-8,0),(-10,3),(-2,2),(-10,4)}
Output:
Boundary points of convex hull are:
(-9, -5) (6, -10) (8, -7) (8, 6) (-7, 8) (-10, 4) (-10, 3)算法
findConvexHull(points, n)
输入：点数，点数。
输出：凸包的角点。
Begin
start := points[0]
for each point i, do
if points[i].x < start.x, then // get the left most point
start := points[i]
done
current := start
add start point to the result set.
define colPts set to store collinear points
while true, do //start an infinite loop
next := points[i]
for all points i except 0th point, do
if points[i] = current, then
skip the next part, go for next iteration
val := cross product of current, next, points[i]
if val > 0, then
next := points[i]
clear the colPts array
else if cal = 0, then
if next is closer to current than points[i], then
add next in the colPts
next := points[i]
else
add points[i] in the colPts
done
add all items in the colPts into the result
if next = start, then
break the loop
insert next into the result
current := next
done
return result
End示例
#include<iostream>
#include<set>
#include<vector>
using namespace std;
struct point { //define points for 2d plane
int x, y;
bool operator==(point p2) {
if(x == p2.x && y == p2.y)
return 1;
return 0;
}
bool operator<(const point &p2)const { //dummy compare function used to sort in set
return true;
}
};
int crossProduct(point a, point b, point c) { //finds the place of c from ab vector
int y1 = a.y - b.y;
int y2 = a.y - c.y;
int x1 = a.x - b.x;
int x2 = a.x - c.x;
return y2*x1 - y1*x2; //if result < 0, c in the left, > 0, c in the right, = 0, a,b,c are collinear
}
int distance(point a, point b, point c) {
int y1 = a.y - b.y;
int y2 = a.y - c.y;
int x1 = a.x - b.x;
int x2 = a.x - c.x;
int item1 = (y1*y1 + x1*x1);
int item2 = (y2*y2 + x2*x2);
if(item1 == item2)
return 0; //when b and c are in same distance from a
else if(item1 < item2)
return -1; //when b is closer to a
return 1; //when c is closer to a
}
set<point>findConvexHull(point points[], int n) {
point start = points[0];
for(int i = 1; i<n; i++) { //find the left most point for starting
if(points[i].x < start.x)
start = points[i];
}
point current = start;
set<point> result; //set is used to avoid entry of duplicate points
result.insert(start);
vector<point> *collinearPoints = new vector<point>;
while(true) {
point nextTarget = points[0];
for(int i = 1; i<n; i++) {
if(points[i] == current) //when selected point is current point, ignore rest part
continue;
int val = crossProduct(current, nextTarget,points[i]);
if(val > 0) { //when ith point is on the left side
nextTarget = points[i];
collinearPoints = new vector<point>; //reset collinear points
}else if(val == 0) { //if three points are collinear
if(distance(current, nextTarget, points[i]) < 0) { //add closer one to collinear list
collinearPoints->push_back(nextTarget);
nextTarget = points[i];
}else{
collinearPoints->push_back(points[i]); //when ith point is closer or same as nextTarget
}
}
}
vector<point>::iterator it;
for(it = collinearPoints->begin(); it != collinearPoints->end(); it++) {
result.insert(*it); //add allpoints in collinear points to result set
}
if(nextTarget == start) //when next point is start it means, the area covered
break;
result.insert(nextTarget);
current = nextTarget;
}
return result;
}
int main() {
point points[] = {{-7,8},{-4,6},{2,6},{6,4},{8,6},{7,-2},{4,-6},{8,-7},{0,0},
{3,-2},{6,-10},{0,-6},{-9,-5},{-8,-2},{-8,0},{-10,3},{-2,2},{-10,4}};
int n = 18;
set<point> result;
result = findConvexHull(points, n);
cout << "Boundary points of convex hull are: "<<endl;
set<point>::iterator it;
for(it = result.begin(); it!=result.end(); it++)
cout << "(" << it->x << ", " <<it->y <<") ";
}输出结果
Boundary points of convex hull are: (-9, -5) (6, -10) (8, -7) (8, 6) (-7, 8) (-10, 4) (-10, 3)