Bobblie Statue Park

Question

This is a Statue Park puzzle.

Rules of Statue Park: (shamelessly stolen from an earlier puzzle by @Deusovi)

• Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.
• Pieces cannot be orthogonally adjacent (though they can touch at a corner).
• All unshaded cells must be (orthogonally) connected.
• Any cells with black circles must be shaded; any cells with white circles must be unshaded.

The piece bank is a set of bobblies, which (long story short) are little crowns with variable number of points. There are 4 no-point bobblies, 3 one-point bobblies, 2 two-point bobblies, and 1 three-point bobblie. I've labeled each with the number of cells they take up. The numbers and backstory have no effect on the puzzle.

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Transcription of puzzle for those who have trouble with images

CSV:

,,,,w,,,,,b
,b,,,b,,,w,w,
,,b,,,,,,w,
w,,,w,,,,b,,w
,,,,b,,,,,
,,,,,b,,,,
b,,w,,,,b,,,b
,b,,,,,,b,,
,w,b,,,b,,,b,w
w,,,,,w,,,w,

There are 4 dominoes, 3 T-shaped tetrominoes, 2 C-shaped hexominoes, and 1 E-shaped octomino in the piece bank.

Answer 1

To start,

take a look at the lower right black cell. It must be part of a 4-piece; any other piece would rub up against a different black dot. And given the placement of two nearby black dots, there's only one way to actually fit it in.

And we can do the same thing again:

Now the next two cells up the diagonal chain have the same deduction! We can't tell exactly how the piece will be placed yet, but it still has to be a 4 piece, or it'll run into a problem with the next black dot.

A big global deduction can be made:

Consider the highlighted dots here. There are 7 red cells; each of them must be part of a separate shape. There are 4 more blue cells; we can only merge one pair of them (either the top two or the bottom two; if we do both the shapes will touch). So this accounts for all 10 of our shapes. In other words, each shape must cover at least one of these cells.

Now, which dots can be 6s? The only available 6 dots are A, D/E, and F: any other placement would either block off the "cubby" of the 6 shape, or brush up against a different known-shaded cell.
And what about 8s? C must have the 8: it can either pair with B, or go vertically by itself.
If it goes vertically, we have a problem: F's spot for the 6 is blocked off, and now we can't fit a piece in B while still pairing E-D. So the 8 goes horizontally, covering B and C.

And now the rest of the pieces fall into place:

E must be a 2. I and J must also be 2s...

And now we need to place one more 4-piece -- if G and H are both ⊣-shaped, then D can't be extended without blocking off the path. So D is the other 4-piece, and the puzzle is solved!