在该问题中,给出了一个多边形,并且还给出了点P。我们需要检查点是在多边形内部还是在多边形外部。
为了求解它,我们将从点P画一条直线。它延伸到无穷大。该线是水平的,或者平行于x轴。
从那条线,我们将计算该线与多边形边相交的次数。当点在多边形内部时,它将与边相交,次数为奇数次,如果将P放置在多边形的任何边上,则它将被剪切偶数次。如果没有一个条件为真,则它在多边形外部。
Input: Points of a polygon {(0, 0), (10, 0), (10, 10), (0, 10)}. And point P (5, 3) to check. Output: Point is inside.
checkInside(Poly, n, p)
输入:多边形的点,多边形的点数,要检查的点p。
输出:当p在多边形内部时为true,否则为false。
Begin if n<3, then return false create a line named exLine from point p to infinity, Slope of the line is 0°. count :=0 and i := 0 repeat create line called side, from point poly[i] to poly[(i+1) mod n] if the side and exLine intersects, then if side and exLine are collinear, then if point p on the side, then return true else return false count := count + 1 i := (i + 1) mod n until i ≠ 0 return true if count is odd End
#include<iostream> using namespace std; struct Point { int x, y; }; struct line { Point p1, p2; }; bool onLine(line l1, Point p) { //check whether p is on the line or not if(p.x <= max(l1.p1.x, l1.p2.x) && p.x <= min(l1.p1.x, l1.p2.x) && (p.y <= max(l1.p1.y, l1.p2.y) && p.y <= min(l1.p1.y, l1.p2.y))) return true; return false; } int direction(Point a, Point b, Point c) { int val = (b.y-a.y)*(c.x-b.x)-(b.x-a.x)*(c.y-b.y); if (val == 0) return 0; //colinear else if(val < 0) return 2; //anti-clockwise direction return 1; //clockwise direction } bool isIntersect(line l1, line l2) { //four direction for two lines and points of other line int dir1 = direction(l1.p1, l1.p2, l2.p1); int dir2 = direction(l1.p1, l1.p2, l2.p2); int dir3 = direction(l2.p1, l2.p2, l1.p1); int dir4 = direction(l2.p1, l2.p2, l1.p2); if(dir1 != dir2 && dir3 != dir4) return true; //they are intersecting if(dir1==0 && onLine(l1, l2.p1)) //when p2 of line2 are on the line1 return true; if(dir2==0 && onLine(l1, l2.p2)) //when p1 of line2 are on the line1 return true; if(dir3==0 && onLine(l2, l1.p1)) //when p2 of line1 are on the line2 return true; if(dir4==0 && onLine(l2, l1.p2)) //when p1 of line1 are on the line2 return true; return false; } bool checkInside(Point poly[], int n, Point p) { if(n < 3) return false; //when polygon has less than 3 edge, it is not polygon line exline = {p, {9999, p.y}}; //create a point at infinity, y is same as point p int count = 0; int i = 0; do { line side = {poly[i], poly[(i+1)%n]}; //forming a line from two consecutive points of poly if(isIntersect(side, exline)) { //if side is intersects exline if(direction(side.p1, p, side.p2) == 0) return onLine(side, p); count++; } i = (i+1)%n; } while(i != 0); return count&1; //when count is odd } int main() { // line polygon = {{{0,0},{10,0}},{{10,0},{10,10}},{{10,10},{0,10}},{{0,10},{0,0}}}; Point polygon[] = {{0, 0}, {10, 0}, {10, 10}, {0, 10}}; Point p = {5, 3}; int n = 4; if(checkInside(polygon, n, p)) cout << "Point is inside."; else cout << "Point is outside."; }
输出结果
Point is inside.