是否在多边形内找到位置C＃

Question

I'm looking for a perfect algorithm in C# which determines whether a location (lat/long) is found within a polygon or not.

Over the Google map I have created different zones in shape of a polygon. Each zone can be represented by set of coordinates. For instance below are the Coordinates for Montego bay (Zone A), Jamaica (longitude,latitude combination):

 -77.9531479,18.4565699,-77.9482339,18.4579542,-77.9443393,18.4615874,-77.9399726,18.4627373,-77.933943,18.4653526,-77.931272,18.4652763,-77.9282138,18.46578,-77.9267121,18.4645891,-77.924619,18.4625118,-77.923131,18.4689837,-77.924646,18.4727285,-77.924112,18.4763919,-77.9255882,18.4793735,-77.927573,18.4833522,-77.9234638,18.4888979,-77.92363,18.492311,-77.914973,18.4951459,-77.918355,18.4971759,-77.972733,18.4985953,-77.916291,18.527569,-77.8997725,18.56178,-77.8962851,18.58578,-77.873843,18.517512,-77.8772736,18.5228297,-77.9253387,18.521568,-77.9531479,18.4565699 

Now lets say we need to find out if Sunset beach Hotels, Montego bay, jamaica whose coordinates are (long/lat): -95.030710, 29.148650 belongs to Zone A or not?

I have tested so many algorithms available around but none of them is working perfectly. It works for some of the hotels, but not for all.

Any help would be appriciated.

Answer 1

I've had success using the Winding Number: http://geomalgorithms.com/a03-_inclusion.html

Step 1: Define the direction the polygon was drawn.

Fundamentally, the polygon is static, and does not have a direction. However, consider the movements you perform when drawing a polygon on a piece of paper. You begin at one position, and move your pen to draw a line for each side of the polygon. If you do not take your pen off the paper, each side of the polygon will either be all drawn in a clockwise direction, or all drawn in a counter-clockwise direction.

We can use this 'direction' to determine how the sides of the polygon 'wind' around the point (clockwise winding or counter-clockwise winding).

With the real polygon in question (which has not been drawn in a direction), it is not important which direction you consider it to have been drawn, but it is important that each side is considered to be drawn in the same direction.

Step 2: Consider each side of the polygon as an individual object with two vertices.

If we know the vertices of the polygon, we can determine the endpoints of each side of the polygon.

Step 3: Determine the direction that each side crosses the point (if the sides do cross the point).

Example using clockwise winding to find whether a polygon surrounds a point :

Define a counter to keep track of winding number.

For each side of the polygon, take the first vertex vertex1 and second vertex vertex2 (these points define the end of each side). If vertex1.x < point.x then side begins to the left of point.x . If vertex2.x > point.x then side ends to the right of point.x . If the side begins to the left AND ends to the right, the line crosses the point.

If the side crosses the point , we need to determine in which direction it crosses. Using a clockwise as positive direction, the side must cross above the point to be considered positive. The line crosses above the point if the y value of the line at the point 'sx position is more than the point 'sy value. Else it crosses below the point.

If it crosses above, increment the counter . If it crosses below, decrement the counter .

NOTE: Notice that if the line is passing from left to right, then positive direction sees the line pass above the point, but if the line goes from right to left (wrapping back under the point) then positive direction sees the line pass below the point. Positive direction is clockwise.

Now do the same thing with the next side of the polygon, where vertex1 would be the previous side's vertex2 , and the new vertex2 is the next corner of the polygon when travelling in a clockwise direction.

Do this for all sides of the polygon.

At the end, you will have a counter with either a positive, negative, or 0 value.

If the value of the counter is 0, the polygon does not contain the point, else, it does.

This is because, a polygon which contains a point will always have an even number of its sides pass the point. Either 2 sides, or 4 sides, etc. Of these sides, half will pass from left to right above the point, and half will pass from right to left below the point.

A result of 0 means either that no sides pass the point (therefore the polygon never crosses the axis of the point), or the polygon passes on one side of the point in two directions (eg above the point left to right, then above the point right to left). Therefore the winding numbers cancel each other out.