O(log n)-time inserts are doable with an augmented binary search tree. To store
order_come | start | end | weight
0 [10, 20] 32
1 [15, 25] 32
2 [5, 30] 32
3 [30, 40] 64
4 [1, 50] 16
5 [1, 60] 16
we have a tree shaped like
25
/ \
/ \
10 50
/ \ / \
5 20 40 60
/ / /
1 15 30 ,
where each number represents the interval from it to its successor. Associated with each tree node are two numbers. The first we call ∆weight, defined to be the weight of the node's interval minus the weight of the node's parent's interval, if extent (otherwise zero). The second we call ∆max, defined to be the maximum weight of an interval corresponding to a descendant of the node, minus the node's weight.
For the above example,
interval | tree node | total weight | ∆weight | ∆max
[1, 5) 1 32 -32 0
[5, 10) 5 64 -32 0
[10, 15) 10 96 32 32
[15, 20) 15 128 32 0
[20, 25) 20 96 0 32
[25, 30) 25 64 64 64
[30, 40) 30 96 64 0
[40, 50) 40 32 16 64
[50, 60) 50 16 -48 80
[60, ∞) 60 0 -16 0
Binary search tree operations almost invariably require rotations. When we rotate a tree like
p c
/ \ / \
c r => l p
/ \ / \
l g g r
we modify
c.∆weight += p.∆weight
g.∆weight += c.∆weight
g.∆weight -= p.∆weight
p.∆weight -= c.∆weight
p.∆max = max(0, g.∆max + g.∆weight, r.∆max + r.∆weight)
c.∆max = max(0, l.∆max + l.∆weight, p.∆max + p.∆weight).
The point of the augmentations is as follows. To find the max weight in the tree, compute r.∆max + r.∆value where r is the root. To increase every weight in a subtree by a given quantity ∂, increase the subtree root's ∆weight by ∂. By changing O(log n) nodes with inclusion-exclusion, we can increase a whole interval. Together, these operations allow us to evaluate an insertion in time O(log n).
To find the total weight of an interval, search for that interval as normal while also adding up the ∆weight values of that interval's ancestors. For example, to find the weight of [15, 30], we look for 15, traversing 25 (∆weight = 64), 10 (∆weight = 32), 20 (∆weight = 0), and 15 (∆weight = 32), for a total weight of 64 + 32 + 0 + 32 = 128.
To find the maximum total weight along a hypothetical interval, we do a modified search something like this. Using another modified search, compute the greatest tree value less than or equal to start (predstart; let predstart = -∞ if start is all tree values are greater than start) and pass it to this maxtotalweight.
maxtotalweight(root, predstart, end):
if root is nil:
return -∞
if end <= root.value:
return maxtotalweight(root.leftchild, predstart, end) + root.∆weight
if predstart > root.value:
return maxtotalweight(root.rightchild, predstart, end) + root.∆weight
lmtw = maxtotalweight1a(root.leftchild, predstart)
rmtw = maxtotalweight1b(root.rightchild, end)
return max(lmtw, 0, rmtw) + root.∆weight
maxtotalweight1a(root, predstart):
if root is nil:
return -∞
if predstart > root.value:
return maxtotalweight1a(root.rightchild, predstart) + root.∆weight
lmtw = maxtotalweight1a(root.leftchild, predstart)
return max(lmtw, 0, root.rightchild.∆max + root.rightchild.∆weight) + root.∆weight
maxtotalweight1b(root, end):
if root is nil:
return -∞
if end <= root.value:
return maxtotalweight1b(root.leftchild, end) + root.∆weight
rmtw = maxtotalweight1b(root.rightchild, end)
return max(root.leftchild.∆max + root.leftchild.∆weight, 0, rmtw) + root.∆weight
We assume that nil has ∆weight = 0 and ∆max = -∞. Sorry for all of the missing details.
