Question

这不是一个与编程相关的问题。但看看你的人是否可以帮助我。

我必须为凸包实现格雷厄姆扫描算法，但问题是我无法找到提供所有信息的伪代码。我发现了一些，但他们留下了一些观点。

感谢。

Answer 1

如果有帮助，请投票：）

#include <iostream> 
#include <stack> 
#include <stdlib.h> 
using namespace std; 

struct Point 
{ 
    int x, y; 
}; 

// A global point needed for  sorting points with reference 
// to  the first point Used in compare function of qsort() 
Point p0; 

// A utility function to find next to top in a stack 
Point nextToTop(stack<Point> &S) 
{ 
    Point p = S.top(); 
    S.pop(); 
    Point res = S.top(); 
    S.push(p); 
    return res; 
} 

// A utility function to swap two points 
int swap(Point &p1, Point &p2) 
{ 
    Point temp = p1; 
    p1 = p2; 
    p2 = temp; 
} 

// A utility function to return square of distance 
// between p1 and p2 
int distSq(Point p1, Point p2) 
{ 
    return (p1.x - p2.x)*(p1.x - p2.x) + 
          (p1.y - p2.y)*(p1.y - p2.y); 
} 

// To find orientation of ordered triplet (p, q, r). 
// The function returns following values 
// 0 --> p, q and r are colinear 
// 1 --> Clockwise 
// 2 --> Counterclockwise 
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;  // colinear 
    return (val > 0)? 1: 2; // clock or counterclock wise 
} 

// A function used by library function qsort() to sort an array of 
// points with respect to the first point 
int compare(const void *vp1, const void *vp2) 
{ 
   Point *p1 = (Point *)vp1; 
   Point *p2 = (Point *)vp2; 

   // Find orientation 
   int o = orientation(p0, *p1, *p2); 
   if (o == 0) 
     return (distSq(p0, *p2) >= distSq(p0, *p1))? -1 : 1; 

   return (o == 2)? -1: 1; 
} 

// Prints convex hull of a set of n points. 
void convexHull(Point points[], int n) 
{ 
   // Find the bottommost point 
   int ymin = points[0].y, min = 0; 
   for (int i = 1; i < n; i++) 
   { 
     int y = points[i].y; 

     // Pick the bottom-most or chose the left 
     // most point in case of tie 
     if ((y < ymin) || (ymin == y && 
         points[i].x < points[min].x)) 
        ymin = points[i].y, min = i; 
   } 

   // Place the bottom-most point at first position 
   swap(points[0], points[min]); 

   // Sort n-1 points with respect to the first point. 
   // A point p1 comes before p2 in sorted output if p2 
   // has larger polar angle (in counterclockwise 
   // direction) than p1 
   p0 = points[0]; 
   qsort(&points[1], n-1, sizeof(Point), compare); 

   // If two or more points make same angle with p0, 
   // Remove all but the one that is farthest from p0 
   // Remember that, in above sorting, our criteria was 
   // to keep the farthest point at the end when more than 
   // one points have same angle. 
   int m = 1; // Initialize size of modified array 
   for (int i=1; i<n; i++) 
   { 
       // Keep removing i while angle of i and i+1 is same 
       // with respect to p0 
       while (i < n-1 && orientation(p0, points[i], 
                                    points[i+1]) == 0) 
          i++; 


       points[m] = points[i]; 
       m++;  // Update size of modified array 
   } 

   // If modified array of points has less than 3 points, 
   // convex hull is not possible 
   if (m < 3) return; 

   // Create an empty stack and push first three points 
   // to it. 
   stack<Point> S; 
   S.push(points[0]); 
   S.push(points[1]); 
   S.push(points[2]); 

   // Process remaining n-3 points 
   for (int i = 3; i < m; i++) 
   { 
      // Keep removing top while the angle formed by 
      // points next-to-top, top, and points[i] makes 
      // a non-left turn 
      while (orientation(nextToTop(S), S.top(), points[i]) != 2) 
         S.pop(); 
      S.push(points[i]); 
   } 

   // Now stack has the output points, print contents of stack 
   while (!S.empty()) 
   { 
       Point p = S.top(); 
       cout << "(" << p.x << ", " << p.y <<")" << endl; 
       S.pop(); 
   } 
} 

// Driver program to test above functions 
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; 
}

Answer 2

这里有我在c ++中实现的graham算法。看看：graham algorithm

Answer 3

我认为你需要这个。 http://aduni.org/courses/algorithms/courseware/psets/Problem_Set_04.doc

用于寻找凸壳的Graham Scan算法

3 个答案: