What is the approach to solve KILLER problem from the May Long Challenge 17? asked 18 May, 18:56

I think my solution is a bit of an overkill, but I'll share it anyway. I heard some people were able to solve this linearly, I'm quite interested so do leave some comments if you're one of them. Short SummaryUse convex hull trick in a persistent segment tree, to minimize quadratic equations. Traverse subtrees using heavylight decomposition. Solution runs in $O(N log^2 N)$. SolutionLet $D_u = h  dep_u$. Define $f$ as the cost of adding the path $(u,v)$, where $v$ is the deeper node: $$ f(u, v) = \sum_{w\in path(u,v)}{(C_vD_w + {C_v}^2)}  H_v $$ $$ = C_v(\frac{D_u(D_u+1)}{2}\frac{D_v(D_v1)}{2})+{C_v}^2(D_uD_v+1)H_v$$ $$ = {D_u}^2\frac{{C_v}}{2} + D_u(\frac{C_v}{2} + {C_v}^2) \frac{C_vD_v(D_v1)}{2} + {C_v}^2(1D_v)  H_v$$ Notice how the only entities with $u$ are $D_u$. Fix $v$, then for every $v$ we have a quadratic equation: $$ f_v(x)= {x}^2\frac{{C_v}}{2} + x(\frac{C_v}{2} + {C_v}^2) \frac{C_vD_v(D_v1)}{2} + {C_v}^2(1D_v)  H_v $$ Notice how quadratic equations are increasing with positive $x$ since $C_v >= 0$. Tree structureWe can use $f_v$ equation to get the value of any node on the path from the root to v. Just plugin $f_v(D_u)$. The red path shows where we can use $f_v$: DP Solution $O(n^2)$Define $dp(v)$ as the answer for the whole subtree rooted at $v$. Before we calculate this, we should shift the parabola $f_v$ by the answer we have for its children: $$f_v(x) := f_v(x) + \sum_{w \in child(v)}{dp(w)}$$ Let $S(u,w)$ be the sum of solutions of the children of $u$ excluding $w$: $$S(u,w) = \sum_{w' \in child(u) \setminus \{w\}}{dp(w')}$$ Then we can compute for $dp(u)$: $$dp(u) = min\{f_u(D_u)\}\cup\{f_v(D_u) + S(u, w) \ \ w \in child(v), v \in subtree(w)\}$$ The answer is $dp(1)$. Traversing each subtree takes $O(N)$ time, and there are $N$ nodes so the total time is $O(N^2)$. This passes subtasks 1 and 2. Convex Hull TrickWe can improve the solution by using the convex hull trick. The goal is instead of iterating through each parabola in the subtree, we want to perform insert/update operations for the log factor. This time is that we are doing this with parabolas and not lines. There are a couple of ways to do this, mine might not be the best. We can do this with any binary search tree, like a splay tree or a multiset. In my case I used ternary search on a segment tree. Before you violently react with "what the heck is a ternary search on a segment tree!", I'll explain how I modeled the segment tree. Notice how every query operation is on an int. Let the segment tree contain the information about the minimum parabola at coordinate $x$, which can be up to the depth of the tree ($10^5$). Query: Go down the segment tree and get the parabola at index $x$. Insert: This is the tricky part. Let $F$ be our set of parabolas (represented by the segment tree). Consider absolute value $F(x)  f_v(x)$. Outside the intersection, the formula has positive value. At the intersection, this formula is $0$. If we want to insert a new parabola $f_v$, we search for the intersection. There can be upto 2 intersections, and that range is where $f_v$ belongs. We can do this in many ways, in my case I did ternary search to find the intersection, since the formula is bitonic and ternary search gives the minimum. Wait, ternary search on a segment tree???We are searching with ints, so instead of splitting $(L,R)$ to $(\frac{2L+R}{3}, \frac{L+2R}{3})$, we can instead split by $(\lfloor{\frac{L+R}{2}}\rfloor, \lfloor{\frac{L+R}{2}}\rfloor+1)$. Yep, ternary search the "binary search" way. (see CF post on Ternary Search hoax) Now that we have the tree, we can solve subtask 3 in $O(N log N)$. With the same $dp(u)$ formula, we can skip traversing the whole subtree and instead use the tree for a point query. Heavylight decomposition + PersistenceFor the convex hull trick to work for a whole tree, we need a way to "pass" a segment tree of parabolas to the parent efficiently. We need to "merge" all parabolas of a node's subtree. But right now, even if we make our segment tree persistent, the best we can do still more than $O(N^2)$ time since merging whole subtrees is costly. This is where HLD comes in. The idea of HLD is to mark a "heavy child" per node in the tree. The heavy child, as its name suggests, is the child with the most descendants. What's useful about the heavy child? In our traversal, we start by "claiming" the heavy child as the current subtree. After that, we traverse the light children and "merge" them one by one. HLD Theorem claims that if we do it this way, the "merging" becomes $O(N log N)$ time total (proof). Truly the magic we need. The code snippet below shows that part:
ComplexityAfter merging, we can perform a query to update the $dp$ function. HLD adds $log N$ factor, while each insert operation is $O(log N)$. For $N$ nodes, the total running time is $O(N log^2 N)$. Btw, this is NOT the official solution editorial, but pardon it for being editoriallike, I can't explain it effectively otherwise. I don't know if my solution matches the intended solution, but I hope that was helpful :) answered 19 May, 19:31
@hikarico Can you suggest any resources to study heavylight decomposition and persistent segment trees?
(22 May, 15:52)

I was thinking I would need something like hikarico's solution. However, what was intended to be an intermediate solution that probably should have TLE'd somehow worked. Start with hikarico's DP O(n^2) solution. Imagine that you are solving this DP bottomup. If we are trying to solve f(x), x will use some node n as the endpoint of its path, where n can be x or a node in the entire subtree of x. If n has a C value of C(n), you can prove the following fact: any other nodes q in x's subtree that have C(q) >= C(n) will NEVER be chosen as the endpoint for any of x's parents. Thus as you go up the tree, you remove such nodes from consideration. So I maintained a vector of nodes under consideration, with a simple O(n) merge + filtering operation. This is O(n^2) worstcase. But the solution passes all the tests in time; I guess the test cases were weak enough for this to succeed. answered 20 May, 05:33

I didn't have enough time to solve it, but it should be possible to solve it by doing a postorder DFS on the tree, while maintaining a kinetic binomial heap over all potential brushes. The binomial heap would allow fast merging on nodes with more than one child and allow you to query the currently cheapest brush. You will have add some extra fields to the data structure to allow fast adding a constant to the heaps that are getting merged. Different from a regular priority queue, you would actually never remove anything from the heap, but always often query the cheapest brush and look at your certificates to update the order of the brushes while moving closer to the root. Kinetic heap: https://en.wikipedia.org/wiki/Kinetic_heap The "time" here is the depth of your current node. As you move closer to the root, some initially expensive brushes, will get cheaper than other options. answered 18 May, 20:31

I can only solve for the case when the tree is just a line. For that case, you can write down the dp problem and the naive iteration will give you O(n^2) but you can apply some convex optimization trick to get O(nlog n). For detail, please check: convex optimization answered 18 May, 22:51

Can you please explain how do you use ternary search for finding intersection points of parabola and convex hull?? answered 10 Nov, 17:37

@hikarico "I did ternary search to find the intersection, since the formula is bitonic and ternary search gives the minimum."  How is the formula given bitonic, it looks like a bimodal function to me with two minimums at 0. Also in the segment tree, what information was stored in each node answered 12 Nov, 19:01
