Editorial : Coloring a Red Black Tree

We show that the problem decomposes into independent subproblems on connected components of black nodes, that each node’s cost is a simple geometric expectation $d/r$, and that a greedy tree DP finds the optimal processing order.

Observation 1 — Independence. Red nodes never change and need no work. Every connected component of black nodes is separated from every other component by red nodes. Solving one component cannot affect another, so the total cost is the sum over all components.

Observation 2 — Geometric cost. When you commit to turning a specific black node $u$ red, each application of the operation picks a uniformly random neighbor. If $u$ has degree $d_u$ and currently $r$ of its neighbors are red, the success probability is $r / d_u$. Independent Bernoulli trials with success probability $p$ need $1/p$ trials in expectation (geometric distribution), so turning $u$ red costs

expected operations. The geometric distribution is memoryless, so interleaving attempts on different nodes offers no advantage: you should pick a node, keep trying until it succeeds, then move on.

Before tackling trees, consider the problem on a path to build intuition.

A contiguous run of $L$ black nodes $b_1, b_2, \ldots, b_L$ with a red node at each end. Every $b_i$ has degree 2.

On a path of length $L$ between two reds, the optimal cost is $2L - 1$. Every node except the last one processed has exactly 1 red neighbor when processed (cost 2); the last node has 2 red neighbors (cost 1). The particular order among the first $L - 1$ nodes doesn’t matter.

On a tree, a high-degree node can accumulate many red neighbors before being processed, making $d_u / r$ much smaller than $d_u / 1$. This makes the order much more impactful and motivates the DP below.

For each connected component of black nodes, pick any node as root and orient edges parent-to-child. Let $r_u$ denote the number of originally-red neighbors of $u$ in the full tree, and $d_u$ its degree in the full tree.

where $i \in \{0, 1\}$:

• $i = 0$: $u$’s parent in the black component tree has not yet been turned red.
• $i = 1$: $u$’s parent has already been turned red (giving $u$ one extra red neighbor).

Let $v_1, \ldots, v_m$ be the black children of $u$. We must choose a subset $S$ of size $k$ to turn red before $u$ (“early” children) and turn the rest red after $u$ (“late” children).

• Each early child $v$ costs $\mathrm{dp}[v][0]$: its parent $u$ is still black.
• Each late child $v$ costs $\mathrm{dp}[v][1]$: its parent $u$ is now red.
• When we process $u$, it has $t = r_u + i + k$ red neighbors, costing $d_u / t$.

Total for a given $k$ and choice of early set $S$:

Rewrite the child costs with a baseline of all-late plus a delta for each early switch:

Define $\delta_v = \mathrm{dp}[v][0] - \mathrm{dp}[v][1] \ge 0$ (having the parent red can only help, so $\mathrm{dp}[v][1] \le \mathrm{dp}[v][0]$). For a fixed $k$, the cheapest early set is the $k$ children with the smallest $\delta_v$.

Sort children by $\delta_v$ ascending. Let $\Sigma_1 = \sum_v \mathrm{dp}[v][1]$ and $\Sigma_k = \sum_{j=1}^{k} \delta_{v_j}$ (prefix sum of sorted deltas). Then:

A leaf of the black component tree has no black children ($m = 0$):

• $\mathrm{dp}[u][0] = d_u / r_u$ if $r_u > 0$, else $+\infty$.
• $\mathrm{dp}[u][1] = d_u / (r_u + 1)$.

Roots use $i = 0$ because they have no parent in the black component tree.

Input: $n = 5$, $s =$ 10010, edges 1–2, 2–3, 3–4, 2–5.

Nodes 1 and 4 are red. Nodes 2, 3, 5 are black and form one connected component.

Root the component at node 2. Children of 2: nodes 3 and 5. Children of 3: none. Children of 5: none.

Leaves:

• $\mathrm{dp}[3][0] = 2 / 1 = 2$, $\quad\mathrm{dp}[3][1] = 2 / 2 = 1$.
• $\mathrm{dp}[5][0] = +\infty$ (no red neighbors), $\quad\mathrm{dp}[5][1] = 1 / 1 = 1$.

Node 2 ($i = 0$, $r_2 = 1$, $d_2 = 3$, children: 3 and 5):

• $\Sigma_1 = \mathrm{dp}[3][1] + \mathrm{dp}[5][1] = 1 + 1 = 2$.
• Sorted deltas: $\delta_3 = 2 - 1 = 1$, $\delta_5 = \infty - 1 = \infty$. Order: $[\delta_3, \delta_5]$.

$\mathrm{dp}[2][0] = 4.5$. Answer $= 4.5$. $\checkmark$

Interpretation: turn node 3 red first (cost 2; it has red neighbor node 4). Then turn node 2 red (cost $3/2 = 1.5$; it now has 2 red neighbors: nodes 1 and 3). Finally turn node 5 red (cost 1; its only neighbor node 2 is now red). Total: $2 + 1.5 + 1 = 4.5$.

Node 5 cannot be turned red before node 2 ($\mathrm{dp}[5][0] = \infty$), so we never consider it as an early child.

• Building the tree and computing $r_u$: $O(n)$.
• DFS: each node is visited once. Sorting children at node $u$ costs $O(m_u \log m_u)$ where $m_u$ is the number of black children. Since $\sum_u m_u = O(n)$, the total sorting cost is $O(n \log n)$.
• Overall: $O(n \log n)$ per test case.

The model solution uses long double for precision and represents $+\infty$ with 1e18. The DFS only recurses into black children (s[v] == '0'), naturally restricting the traversal to one connected component. The outer loop for i = 1 to n finds unvisited black nodes as component roots.

The second dimension of the DP ($i \in \{0, 1\}$) is computed in the same DFS call, reusing the same sorted delta array for both values of $i$.