## program solution

// A C++ program to check if a given point lies inside a given polygon

#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);
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);
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);
isInside(polygon3, n, p)? cout << "Yes \n": cout << "No \n";

return 0;
}

No

Yes

Yes

Yes

No

No