Flatten transparent pixels to reduce palette size

[Muccarini]

When creating frames from RGBA data with Frame::from_rgba_speed(), all transparent pixels (alpha = 0) are now flattened to use the same RGB values. This ensures transparent pixels count as a single color entry instead of multiple entries in the palette.

Benefit: Reduces palette size and helps avoid unnecessary NeuQuant quantization when the image has ≤256 unique colors.

The BTreeSet<(u8, u8, u8, u8)> now counts: opaque colors + 1 transparent color.

like suggested in #110

One concern I have is around handlng of "transparent". For NeuQuant, it looks to me like if there's multiple colours that have 0 alpha, only one gets selected as transparent, but independently NeuQuant will build an RGBA palette for all the given colours, and of multiple colours with 0 alpha, only one will be mapped to transparent. The others will become visible.
. . . I suggest the pre-pass converts all transparent pixels to explicitly the same RGBA value to avoid this issue.

PS. to avoid .unwrap() i inserted a redundant Some() check. my rust knowledge is limited, any suggestion is appreciated.

[fintelia]

I'd probably suggest either this:

if pix[3] != 0 {
    pix[3] = 0xFF;
} else {
    match transparent {
        ...
    }
}

Or this:

if pix[3] != 0 {
    pix[3] = 0xFF;
    continue;
}

match transparent {
    ...
}

[197g]

I'd use if let instead of match here since we only need contents from the Some-branch but follow #224 (comment) otherwise.

Suggested change

	} else if transparent.is_none() {
	transparent = Some([pix[0], pix[1], pix[2], pix[3]]);
	} else if let Some([r, g, b, a]) = transparent {
	pix[0] = r;
	pix[1] = g;
	pix[2] = b;
	pix[3] = a;
	continue;
	}
	if let Some([r, g, b, a]) = transparent {
	pix[0] = r;
	pix[1] = g;
	pix[2] = b;
	pix[3] = a;
	} else {
	transparent = Some([pix[0], pix[1], pix[2], pix[3]]);
	}