问注满二维网格

EN

a,X,Y=input()
M=len(a)
N=len(a[0])
def R(x,y):
 if~0<x<M and~0<y<N and a[x][y]:a[x][y]=0;R(x-1,y);R(x+1,y);R(x,y-1);R(x,y+1)
R(X,Y)
print'\n'.join(map(str,a))

[©g_#®ć©d®IgIнg‚‹«PisD®нèD®θèi2®θǝ®нǝs1Ý‚D(«ε®+}ìë\s

[                  # Start an infinite loop:
 ©                 #  Store the list of coordinates in variable `®` (without
                   #  popping)
                   #  (which is the first implicit input in the first iteration)
  g_               #  If it's empty:
    #              #   Stop the infinite loop
  ©ć               #  Extract head of `®`; pop and remainder-list and first
                   #  coordinate separated to the stack
    ©              #  Store this coordinate as new variable `®` (without popping)
     d             #  Check [x>=0,y>=0]
       Ig          #  Push the amount of rows of the input-matrix
         Iнg       #  Push the amount of columns of the input-matrix
            ‚      #  Pair them together
      ®      ‹     #  Check [x<rowCount,y<columnCount]
              «    #  Merge the two lists together
               Pi  #  Pop, and if all were truthy (so [x,y] is within bounds):
 s                 #   Swap so the matrix is at the top
  D                #   Duplicate this matrix
   ®нè             #   Pop and get the x'th (of coordinate `®`) row of the matrix
      D            #   Duplicate this row
       ®θè         #   Pop and get the y'th (of coordinate `®`) cell of this row
          i        #   If this value is 1:
           2®θǝ    #    Insert a 2 at this y'th cell of the row
               ®нǝ #    Insert this modified x'th row into the matrix
 s                 #    Swap so the list of coordinates is at the top again
  1Ý               #    Push [0,1]
    ‚             #    Bifurcate (Duplicate & Reverse copy) and pair:
                   #    [[0,1],[1,0]]
      D(«          #    Duplicate, negate, merge: [[0,1],[1,0],[0,-1],[-1,0]]
         ε®+}      #    Add coordinate `®` to each pair:
                   #    [[x,y+1],[x+1,y],[x,y-1],[x-1,y]]
             ì     #    Prepend-merge it at the front of the coordinates-list
          ë        #   Else: the value was 0 or 2 instead:
           \       #    Discard the duplicated row
            s      #    And swap so the coordinates list is at the top again
]                  # Close the open if-statements and infinite loop
 \                 # Discard the empty coordinates list
                   # (after which the matrix is output implicitly as result)

int f(int[][]m,int x,int y){try{if(m[x][y]==0)f(m,x+f(m,x,y+f(m,x-f(m,x,y+--m[x][y]),y)),y);}finally{return 1;}}

// Recursive method with integer matrix & x,y parameters and integer return-type
int f(int[][]m,int x,int y){
  try{if(m[x][y]==0)   //  If the current cell-value is 0:
            --m[x][y]  //   Decrement the current cell-value to -1
           f(m,x,y+...)//   Recursive call west
          f(m,x-...,y) //   Recursive call north
         f(m,x,y+...)  //   Recursive call east
        f(m,x+...,y);  //   Recursive call south
  }finally{            //  Catch and swallow any ArrayIndexOutOfBoundsExceptions
                       //  (shorter than manual if-checks)
           return 1;}  //  Always result in 1, so we use it to our advantage to
                       //  save bytes here by nesting, instead of having four
                       //  loose calls to each direction)