How to improve time complexity of this code piece? - ruby

Question

I am having a hard time trying to down down the time. If this can be done in O(n^2) thatd be awesome.
def printDistance(file)
line = file.gets
if line == nil then return end
sz, sx, sy, ex, ey = line.split(/\s/)
#counter = 0
while line = file.gets do
if line[0...4] == "path"
else
x, y, ds, w = line.split(/\s/,4)
x = x.to_i
y = y.to_i
matrix= Array.new
later = Array.new
tmp = Array.new
processing = Array.new
if matrix[y].class != Array
matrix[y] = Array.new
end
if ds == "" #this cell has NO way to get to
matrix[y][x] = nil
else
matrix[y][x] = ds
end
end
end
for y in 0...matrix.length
processing[y] = Array.new
tmp[y] = Array.new
later[y] = Array.new
end
printing = Array.new
counter = 0
sy = sy.to_i
sx = sx.to_i
processing[sy][sx] = matrix[sy][sx]
matrix[sy][sx] = nil
puts "#{counter},(#{sx},#{sy})" #first one
counter += 1
loop do
print "#{counter},"
counter += 1
for y in 0...processing.length
for x in 0...processing[y].length
if processing[y][x].class != nil
tmp[y][x] = processing[y][x]
dirArr = tmp[y][x].to_s.split(//)
dirArr.each { |c|
if c == "u"
newY = y - 1
newX = x
elsif c == "d"
newY = y + 1
newX = x
elsif c == "l"
newY = y
newX = x - 1
else #c == r
newY = y
newX = x + 1
end
if matrix[newY][newX] != nil
tmpStr = "(#{newX},#{newY})"
printing.unshift(tmpStr)
later[newY][newX] = matrix[newY][newX]
matrix[newY][newX] = nil
end
}
end
end
end
printing.sort!
for i in 0...printing.length
print printing[i]
if i < printing.length - 1
print ","
end
end
printing = [] #resetting
puts
for i in 0...later.length
for j in 0...later[i].length
if later[i][j] != nil
processing[i][j] = later[i][j]
end
end
end
break if NotEmpty(matrix) == false
end
end
def NotEmpty (a)
for i in 0...a.length
for j in 0...a[i].length
if a[i][j] != nil
return true
end
end
end
return false
end
Basically this code reads in a file and places it in a 2-d array representing a maze, then based on maze's starting point it will perform a BFS to put all cells in order from closest to the start to the farthest, then print everything out
This code is now O(n^3) but I am trying to shrink it to O(n^2), is there anyway I can traverse a 2-d array and keeping track of the x and y values without using two forloops?
Any help is appreciated!! Thanks!