检查点是否位于多边形内？

Question

我已经看到了在 gfg 中检查点位于多边形内部或外部的概念的实现 但我需要通过如果点位于多边形的边缘或顶点上来更新它，那么不考虑在里面。 这是下面的代码。我必须在哪里更新？

    // A C++ program to check if a given point lies inside a given polygon 
// Refer https://www.geeksforgeeks.org/check-if-two-given-line-segments-intersect/ 
// for explanation of functions onSegment(), orientation() and doIntersect() 
#include <iostream> 
using namespace std; 

// Define Infinite (Using INT_MAX caused overflow problems) 
#define INF 10000 

struct Point 
{ 
    int x; 
    int y; 
}; 

// Given three colinear points p, q, r, the function checks if 
// point q lies on line segment 'pr' 
bool onSegment(Point p, Point q, Point r) 
{ 
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) && 
            q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y)) 
        return true; 
    return false; 
} 

// 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 
} 

// The function that returns true if line segment 'p1q1' 
// and 'p2q2' intersect. 
bool doIntersect(Point p1, Point q1, Point p2, Point q2) 
{ 
    // Find the four orientations needed for general and 
    // special cases 
    int o1 = orientation(p1, q1, p2); 
    int o2 = orientation(p1, q1, q2); 
    int o3 = orientation(p2, q2, p1); 
    int o4 = orientation(p2, q2, q1); 

    // General case 
    if (o1 != o2 && o3 != o4) 
        return true; 

    // Special Cases 
    // p1, q1 and p2 are colinear and p2 lies on segment p1q1 
    if (o1 == 0 && onSegment(p1, p2, q1)) return true; 

    // p1, q1 and p2 are colinear and q2 lies on segment p1q1 
    if (o2 == 0 && onSegment(p1, q2, q1)) return true; 

    // p2, q2 and p1 are colinear and p1 lies on segment p2q2 
    if (o3 == 0 && onSegment(p2, p1, q2)) return true; 

    // p2, q2 and q1 are colinear and q1 lies on segment p2q2 
    if (o4 == 0 && onSegment(p2, q1, q2)) return true; 

    return false; // Doesn't fall in any of the above cases 
} 

// Returns true if the point p lies inside the polygon[] with n vertices 
bool isInside(Point polygon[], int n, Point p) 
{ 
    // There must be at least 3 vertices in polygon[] 
    if (n < 3) return false; 

    // Create a point for line segment from p to infinite 
    Point extreme = {INF, p.y}; 

    // Count intersections of the above line with sides of polygon 
    int count = 0, i = 0; 
    do
    { 
        int next = (i+1)%n; 

        // Check if the line segment from 'p' to 'extreme' intersects 
        // with the line segment from 'polygon[i]' to 'polygon[next]' 
        if (doIntersect(polygon[i], polygon[next], p, extreme)) 
        { 
            // If the point 'p' is colinear with line segment 'i-next', 
            // then check if it lies on segment. If it lies, return true, 
            // otherwise false 
            if (orientation(polygon[i], p, polygon[next]) == 0) 
            return onSegment(polygon[i], p, polygon[next]); 

            count++; 
        } 
        i = next; 
    } while (i != 0); 

    // Return true if count is odd, false otherwise 
    return count&1; // Same as (count%2 == 1) 
} 

// Driver program to test above functions 
int main() 
{ 
    Point polygon1[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}}; 
    int n = sizeof(polygon1)/sizeof(polygon1[0]); 
    Point p = {20, 20}; 
    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n"; 

    p = {5, 5}; 
    isInside(polygon1, n, p)? cout << "Yes \n": cout << "No \n"; 

    Point polygon2[] = {{0, 0}, {5, 5}, {5, 0}}; 
    p = {3, 3}; 
    n = sizeof(polygon2)/sizeof(polygon2[0]); 
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n"; 

    p = {5, 1}; 
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n"; 

    p = {8, 1}; 
    isInside(polygon2, n, p)? cout << "Yes \n": cout << "No \n"; 

    Point polygon3[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}}; 
    p = {-1,10}; 
    n = sizeof(polygon3)/sizeof(polygon3[0]); 
    isInside(polygon3, n, p)? cout << "Yes \n": cout << "No \n"; 

    return 0; 
}

它会正常工作。仅当点位于多边形的边界时才需要更新，将被认为不在多边形内部（与外部相同）。 在此先感谢

Answer 1

正如在 isInside 函数内的 cmets 中解释的那样：

    // If the point 'p' is colinear with line segment 'i-next', 
    // then check if it lies on segment. If it lies, return true, 
    // otherwise false 
    if (orientation(polygon[i], p, polygon[next]) == 0) 
    return onSegment(polygon[i], p, polygon[next]); 

如果你想在要定义为外部的线段上点，那么你只需要像这样反转这个返回值。

    return !onSegment(polygon[i], p, polygon[next]); 

Answer 2

这个函数检查点是否在段上。

bool onSegment(Point p, Point q, Point r) 
{ 
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) && 
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y)) 
    return true; 
  return false; 
} 

所以你需要反转你调用这个函数的逻辑

 if (doIntersect(polygon[i], polygon[next], p, extreme)) 
    { 
        // If the point 'p' is colinear with line segment 'i-next', 
        // then check if it lies on segment. If it lies, return true, 
        // otherwise false 
        if (orientation(polygon[i], p, polygon[next]) == 0) 
        return !onSegment(polygon[i], p, polygon[next]); 

        count++; 
    } 

或者你可以改变函数本身

 bool notOnSegment(Point p, Point q, Point r) 
{ 
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) && 
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y)) 
    return false; 
  return true; 
} 

并调用它

     if (doIntersect(polygon[i], polygon[next], p, extreme)) 
    { 
        // If the point 'p' is colinear with line segment 'i-next', 
        // then check if it lies on segment. If it lies, return true, 
        // otherwise false 
        if (orientation(polygon[i], p, polygon[next]) == 0) 
        return notOnSegment(polygon[i], p, polygon[next]); 

        count++; 
    }