Fillomino Connection is a puzzle type original to this blog to the best of my knowledge (although I admit I didn’t really search, and I would be surprised if it’s really the first time). It is based on nikoli’s Fillomino.
Like in a Fillomino puzzle, some cells are filled by numbers. Some cells are also shaded, meaning that they must be connected. Same-colored cells must be connected to each other by same-colored path.
Just like in a regular Fillomino, a cell interacts with its four adjacent neighbors. If two cells have the same number, then they are in the same region, otherwise they are in different regions. The size of a region must be equal to the number in the region.
Now, with shaded cells, all regions that contain a shaded cell are shaded. A region may not consist of a shaded clue and an unshaded clue, or in case there are two or more colors, a region also may not consist of differently colored clues. All regions that belong to the same color forms a contiguous region of its color; this means starting from any square of a color other than white, it is possible by passing the regions of the same color to visit any other square of the same color.
Fill each cell with a number and shade the regions whose at least one shaded number belongs to it with the same color such that each number indicates the size of their region and all regions of a certain color are connected.
Walkthrough of the example
Let’s begin with the lower-left corner. R4C2 cannot be 3, otherwise a region of a shaded clue and an unshaded clue is formed. So R4C1 doesn’t extend to R4C2, thus it must extend to R3C1.
Note that the gray region sums up to an area of 5, and that the least region containing both R1C2 and R4C3 has an area of 5 squares, so the two regions must be effective; that is, none of them goes outside the 4×2 rectangle of R1C2-R4C3. Thus, the 3 must extend upward to R3C3 and R2C3. It’s simple to fill that R2C1 is also a 3, forming a region. Since R1C2 must extend to either R1C3 or R2C2, then R1C1 is isolated, thus it must be a 1.
If R1C2 extends to R2C2, the remaining area of R2C3-R2C4 cannot be filled. The only possibility is that they are both 2, and thus R2C2 extends to R2C3 and R2C4, contradiction. (Another possible way is that if R1C2 extends to R2C2, R1C3 is isolated and thus must be 1, contradicting with R1C4.) Thus R1C2 must extend to R1C3, and it’s easy to fill in the rest by simple trial and error.