Approach #1: Reverse Time and Union-Find [Accepted]
The problem is about knowing information about the connected components of a graph as we cut vertices. In particular, we'll like to know the size of the "roof" (component touching the top edge) between each cut. Here, a cut refers to the erasure of a vertex.
As we may know, a useful data structure for joining connected components is a disjoint set union structure. The key idea in this problem is that we can use this structure if we work in reverse: instead of looking at the graph as a series of sequential cuts, we'll look at the graph after all the cuts, and reverse each cut.
We'll modify our typical disjoint-set-union structure to include a
dsu.size operation, that tells us the size of this component. The way we do this is whenever we make a component point to a new parent, we'll also send it's size to that parent.
We'll also include
dsu.top, which tells us the size of the "roof", or the component connected to the top edge. We use an ephemeral "source" node with label
R * C where all nodes on the top edge (with row number
0) are connected to the source node.
For more information on DSU, please look at Approach #2 in the article here.
Next, we'll introduce
A, the grid after all the cuts have happened, and initialize our disjoint union structure on the graph induced by
A (nodes are grid squares with a brick; edges between 4-directionally adjacent nodes).
After, if we get an cut at
(r, c) but the original
grid[r][c] was always
0, then we couldn't have had a meaningful cut - the number of dropped bricks is
Otherwise, we'll look at the size of the new roof after adding this brick at
(r, c), and compare them to find the number of dropped bricks.
Since we were working in reverse time order, we should reverse our working answer to arrive at our final answer.
Time Complexity: , where is the number of grid squares, is the length of
hits, and is the Inverse-Ackermann function.
Space Complexity: .
Analysis written by: @awice.