Question

我已经看到了在gfg中检查点位于多边形内部还是外部的概念的实现但是我需要通过更新点是否位于多边形的边缘或顶点上来进行更新，而不是将其内部考虑在内。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; 
}

它将正常工作。仅当点位于多边形的边界内时才需要更新，将被视为不在多边形内部（与外部相同）。预先感谢

Answer 1

如isInside函数内的注释所述：

    // 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]);

如果要指向要定义为外部的线段，则只需反转此返回值即可。

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

Answer 2

此功能正在检查点是否在线段上。

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; 
}

因此，您需要从调用此函数的过程中逆转逻辑

 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++; 
    }

或者您可以更改功能本身

 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

     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++; 
    }

检查点是否位于多边形内？

2 个答案: