05AB1E, 85 bytes

2FINi(}Θ©˜!ƶIgä®*Δ2Fø0δ.ø}2Fø€ü3}®*εεÅsyøÅs«à]+ÐU˜ê©¦vyX®¦y†AS:¾1:'a0:4FJTx:€Søí}S2¢: 

Try it online or verify all test cases.

TODO: Will add an explanation later today. I'm first going to lunch and have a meeting at work.

As a short summary for now:

1. Flood-fill the individual islands, which I've done in 05AB1E before for this challenge (which actually contained a bug that I discovered thanks to the final test case here);
2. Loop over the islands:
   1. Convert the matrix to something usable, so a similar approach can be used as in the linked challenge (converts 0s to 1s; the current island to 0s; and the other islands to lowercase letters).
   2. Actually use that approach to calculate the counts of each island
   3. Replace the islands with these counts.

05AB1E, 85 81 bytes

2FINi(}Θ©˜ƶIgäΔ2Fø0δ.ø}2Fø€ü3}®*εεÅsyøÅs«à]+ÐU˜ê¦©vyX®y†AS:¾1:'a0:4FJTx:€Søí}S2¢: 

Try it online or verify all test cases.

Step 1: Flood-fill the individual islands. I've done this before in 05AB1E for this challenge (which actually contained a bug that I discovered thanks to the final test case here):

2F # Loop 2 times (for each 1/-1 type of piece individually): I # Push the input-matrix Ni } # If the 0-based loop-index is 1: ( # Negate all values Θ # Check which values are equal to 1 (maps 1→1; 0→0; -1→0) © # Store this in variable `®` (without popping) ˜ # Flatten it to a list ƶ # Multiply each value by its 1-based index ä # Convert it back to the matrix, Ig # using the input-length amount of rows Δ # Loop until it no longer changes to flood-fill: 2Fø0δ.ø} # Add a border of 0s around the matrix: 2F } # Loop 2 times: ø # Zip/transpose; swapping rows/columns δ # Map over each row: 0 .ø # Add a leading/trailing 0 2Fø€ü3} # Convert it into overlapping 3x3 blocks: 2F } # Loop 2 times again: ø # Zip/transpose; swapping rows/columns € # Map over each inner list: ü3 # Convert it to a list of overlapping triplets ®* # Multiply each 3x3 block by the value in matrix `®` # (so the 0s remain 0s) εεÅsyøÅs«à # Get the largest value from the horizontal/vertical cross of # each 3x3 block: εε # Nested map over each 3x3 block: Ås # Pop and push its middle row y # Push the 3x3 block again ø # Zip/transpose; swapping rows/columns Ås # Pop and push its middle rows as well (the middle column) « # Merge the middle row and column together to a single list à # Pop and push its maximum ] # Close the nested maps, flood-fill loop, and outer loop + # Add the values in the two matrices together 

Try just step 1 online.

Step 2: Loop over the individual islands, and convert all 0s to 1s; the current island to 0s; and all other islands to lowercase letters, as preparation for the next step:

Ð # Triplicate the resulting matrix U # Pop one, and save it in variable `X` ˜ # Flatten another one ê # Sort and uniquify the values ¦ # Remove the leading 0 © # Store these island-values in variable `®` (without popping) v # Loop over each island-value `y`: X # Push matrix `X` ® # Push list `®` y† # Filter the current island to the front AS # Push the lowercase alphabet as list : # Replace the island-values with letters ¾1: # Replace all 0s with 1s 'a0: # Replace all "a" (the current island) with 0s 

Try the first 2 steps online (with some added debug-lines).

Step 3: Use the same method as in the linked challenge to get the liberty-counts of each island:

 4F # Loop 4 times: J # Join the inner rows together to strings T # Push 10 x # Double it (without popping): 20 : # Replace all "10" with "20" €S # Convert each row back to a list again øí # Rotate the matrix 90 degrees clockwise: ø # Zip/transpose; swapping rows/columns í # Reverse each inner row } # Close the loop S # Convert the matrix to a flattened list of characters 2¢ # Count how many 2s are in this list 

Try the first 3 steps online (again with added debug-lines).

Step 4: Convert the current island to this count in the matrix:

 # The matrix from the triplicate of step 2 is still on the stack y # Push island-value `y` from the still open loop of step 2 : # Replace all of these islands with its count from step 3 # (after which the resulting matrix is output implicitly)