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[learning notes] radical thinking question

~~Violent data structures~~

Consider the whole block
Because the fragmentation is very obvious , Just refactor directly
When merging the two parties , We cannot superimpose the marks of one party directly on the marks of another party
So we need to create a new point , Maintain the parties it represents and the cumulative value
Obviously, we are building a tree
When recruiting , In fact, it is the accumulation of all points that become the current party
Obviously, it can be handled in reverse order
Obviously, you can consider each block separately
So the spatial complexity $O (n)$
The complexity of time is obviously $O(n\sqrt{n})$