Zoj1081 Problem Solving report

The general meaning of the question: Create a polygon for several points, and then give some points to determine whether the points are inside or outside the polygon.

A template question... I can't do mathematics. I can't write it myself. I understand the general idea. I searched the internet. There are many methods, one of which is to make the rays parallel to the X axis, then, determine the number of intersections. The odd number is inside and the even number is outside. note that the X-ray does not count when it passes through the vertices of a polygon. ah, it seems that the mathematics is too weak. the Code is as follows:

# Include <stdio. h>
# Include <math. h>
# Define max (A, B) (a)> (B )? (A) (B ))
# Define min (A, B) (a) <(B )? (A) (B ))
# Define infinity 1e9
# Define ESP 1e-7
# Define max_n 100
Struct point
{
Double X, Y;
};
Struct linesegment
{
Point pt1, pt2;
};
// Calculate the cross multiplication | p1p0 | × | p2p0 |
Double multiply (point P1, point P2, point P0)
{
Return (p1.x-Snapshot X) * (p2.y-Snapshot y)-(p2.x-Snapshot X) * (p1.y-Snapshot y ));
}

Bool isonline (point, linesegment line)
{
Return (FABS (multiply (line. pt1, line. pt2, point) <ESP )&&
(Point. x-line.pt1.x) * (point. x-line.pt2.x) <= 0 )&&
(Point. y-line.pt1.y) * (point. y-line.pt2.y) <= 0 ));
}
// Determine the intersection of line segments
Bool intersect (linesegment L1, linesegment l2)
{
Return (max (l1.pt1. X, l1.pt2. X)> = min (l2.pt1. X, l2.pt2. X ))&&
(Max (l2.pt1. X, l2.pt2. X)> = min (l1.pt1. X, l1.pt2. X ))&&
(Max (l1.pt1. Y, l1.pt2. Y)> = min (l2.pt1. Y, l2.pt2. y ))&&
(Max (l2.pt1. Y, l2.pt2. Y)> = min (l1.pt1. Y, l1.pt2. y)
(Multiply (l2.pt1, l1.pt2, l1.pt1) * multiply (l1.pt2, l2.pt2, l1.pt1)> = 0 )&&
(Multiply (l1.pt1, l2.pt2, l2.pt1) * multiply (l2.pt2, l1.pt2, l2.pt1)> = 0 ));
}

// Points in the Polygon
Int inpolygon (point polygon [], int N, point)
{
If (n = 1)
Return (FABS (polygon [0]. x-point.x) <ESP) & (FABS (polygon [0]. y-point.y) <ESP ));
Else if (n = 2)
{
Linesegment side;
Side. pt1 = polygon [0];
Side. pt2 = polygon [1];
Return isonline (point, side );
}
Int COUNT = 0, I;
Linesegment line;
Line. pt1 = point;
Line. pt2.y = point. Y;
Line. pt2.x =-infinity;

For (I = 0; I <n; I ++)
{
Linesegment side;
Side. pt1 = polygon [I];
Side. pt2 = polygon [(I + 1) % N];

If (isonline (point, side ))
Return true;

// If the side parallel X axis is not considered
If (FABS (side. pt1.y-side. pt2.y) <ESP)
Continue;
// 4. An endpoint is on a straight line
If (isonline (side. pt1, line ))
{
If (side. pt1.y> side. pt2.y)
Count ++;
}
Else if (isonline (side. pt2, line ))
{
If (side. pt2.y> side. pt1.y)
Count ++;
}
// Intersection
Else if (intersect (line, side ))
Count ++;
}

Return count % 2;
}

Int main ()
{
// Freopen ("in.txt", "r", stdin );
Int n, m, casenum = 0, I;
Point polygon [max_n];
Point P;
While (scanf ("% d", & N)
{
Casenum ++;
If (casenum> 1)
Printf ("\ n ");
Printf ("problem % d: \ n", casenum );
Scanf ("% d", & M );
For (I = 0; I <n; I ++)
Scanf ("% lf", & polygon [I]. X, & polygon [I]. y );

For (I = 0; I <m; I ++)
{
Scanf ("% lf", & P. X, & P. y );
If (inpolygon (polygon, n, p ))
Printf ("within \ n ");
Else
Printf ("outside \ n ");
}
}
Return 0;
}

Zoj1081 Problem Solving report