Check for closed path in numerical heightmap

Question

Hi all, I'm working on a game, and I have the necessity to check a closed path in a given numerical heightmap: The server and the client use this heightmap to set the right coords to move etc... Now, when an user walks on a "special" tile, it lights... My problem is: When an user, walking on these tiles creates a closed path with empty tiles in it, the server should automatically fill the tiles in this path...

It should do something like this: http://www.youtube.com/watch?v=kAVUNE2NTUQ - 1:32

I'm sure I've to use some maths here or there, but I dunno how... I could do a "for" cycle, but it would be too long, and the problem is that the server needs to do the cycle every time an user walks... Thanks in advance for your answers, hope someone could help me.

PS: I'm using C#

EDIT: When an user walks on a tile, the server automatically replaces the heightmap[X, Y] with an integer that represents the color of the user

Answer 1

"the problem is that the server needs to do the cycle every time an user walks"...

You can eliminate the cycling over walked tiles by assigning each tile in the game a unique integer, having an array with the number of tiles, and then for each step mark this tile in the array. Of course, also check whether the tile has been walked before (i.e. whether it's marked in the array), and if it has, you have a loop (maybe of zero area). This approach has no cycling over walked tiles, etc, but just a single look-up for each step.

Answer 2

Let's assume you have rectangular field with upper left corner in (0, 0) and bottom right in (M, N). Then you can use the following pseudocode fill algorithm:

find the upper and bottom cells of the path (yTop, yBottom)

for (int y = yTop; y != yBottom; ++y)
{
    find leftmost and rightmost path cells (xLeft, xRight) for that y
    bool isInside = true;
    for (int x = xLeft+1; x<xRight; ++x)
    {
        if ((x, y) is path cell)
            isInside = !isInside;
        if (isInside)
            fill(x, y);
    }
}