Editorial for DMOPC '21 Contest 8 P5 - Tree Building - Olympiads Online Judge

Editorial for DMOPC '21 Contest 8 P5 - Tree Building

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.
Submitting an official solution before solving the problem yourself is a bannable offence.

Author: zhouzixiang2004

Subtask 1

Hint

Which sequences of $N$ numbers can be the degree sequence of a tree with $N$ nodes?

Solution

For subtask 1, note that any sequence of integers $d_1, d_2, \dots, d_N$ with $d_i \ge 1$ for all $i$ and $\sum_{i=1}^N d_i = 2N-2$ can be the degree sequence of a tree with $N$ nodes.

Proof

Clearly, these conditions are necessary. To prove they are sufficient, induct on $N$ with the $N = 2$ case being clear. As $\sum_{i=1}^N d_i = 2N-2$ , there exists indices $i$ and $j$ with $d_i \ge 2$ and $d_j = 1$ , so we can subtract $1$ from $d_i$ , delete $d_j$ , inductively construct a tree for the remaining sequence, and attach a leaf to node $i$ labelled $j$ .

We can now perform a 2D knapsack dynamic programming, where $dp[i][j]$ represents the minimum $\sum_{k=1}^i a_{d_k}$ if $\sum_{k=1}^i d_k = j$ , with $K$ transitions at each state.

Time Complexity: $\mathcal{O}(N^2K)$

Subtask 2

Hint

The subtask 1 solution performs the same transitions $N$ times. There is a technique, similar to fast exponentiation, that reuses previously computed results so that we only need to perform the transitions $\mathcal{O}(\log N)$ times.

Solution

The subtask 1 solution transitions only from $dp[i][\bullet]$ to $dp[i+1][\bullet]$ by looping through all possible values of $d_{i+1}$ (with additional cost $a_{d_{i+1}}$ ). However, we can also transition directly from $dp[i][\bullet]$ to $dp[2i][\bullet]$ by looping through all possible values of $d_{i+1} + d_{i+2} + \dots + d_{2i}$ (with additional cost $dp[i][d_{i+1} + d_{i+2} + \dots + d_{2i}]$ ). This allows us to compute $dp[N][\bullet]$ using only $\mathcal{O}(\log N)$ transitions.

This technique is known as the " $\times 2+1$ " trick. If we view the transitions as performing min-plus convolutions (and $dp[N][\bullet]$ as a min-plus convolution to the power of $N$ ), the similarities between the $\times 2+1$ trick and fast exponentiation may become more clear.

Time Complexity: $\mathcal{O}(N^2\log N)$

Subtask 3

Hint

Find a dynamic programming approach that only uses $\mathcal{O}(N)$ states.

Hint 2

The subtask 1 approach may be misleading. Try to find a recurrence that works more directly with the tree structure.

Solution

Consider a group of $j$ leaves with the same parent. If we delete all of these leaves, we have a smaller tree, and the degree of the parent node is $1$ which gets replaced by a degree of $j+1$ . This leads to a new dynamic programming formulation, where we let $dp[i]$ be the minimum cost of a tree with $i$ nodes, with transitions of the form $\displaystyle dp[i] = \min_{j=1}^{K-1} dp[i-j]+a[j+1]+(j-1)a[1]$ Time Complexity: $\mathcal{O}(NK)$

Subtask 4

Hint

View the subtask 3 solution as an instance of unbounded knapsack with $K-1$ items. When $N$ is really large, what items should we take?

Solution

We can write the subtask 3 recurrence in the form $\displaystyle dp[i] = \min_{j=1}^{K-1} dp[i-j]+b[j]$ where $b[j] = a[j+1]+(j-1)a[1]$ for $1 \le j \le K-1$ , which is an instance of unbounded knapsack.

Intuitively, when $N$ is really large, we should repeatedly take the $j$ with the smallest $\frac{b[j]}{j}$ as this minimizes the cost per unit of "weight." In fact, this is true whenever $N > K^2$ .

Proof

Suppose $N > K^2$ and we must never use this minimal $j$ . As each item occupies a "weight" of at most $K-1$ , we must have used $k > K > j$ items, say $u_1, u_2, \dots, u_k$ to form $dp[N]$ (possibly with duplicates). It is well known that in any set of $k \ge j$ numbers, some nonempty subset of them sums to a multiple of $j$ (Proof: Pigeonhole on the $k+1$ prefix sums modulo $j$ and take their difference). Applying this lemma to $u_1, u_2, \dots, u_k$ , some nonempty subset of them sum to a multiple of $j$ . We can replace this subset with copies of item $j$ , which doesn't make the result worse since $\frac{b[j]}{j}$ is minimal, giving a contradiction.

How tight is this bound?

It is tight up to terms linear in $K$ . We may have a test case where $b[K-2]$ and $b[K-1]$ are the only $b[i]$ that aren't too big to be useful, with $\frac{b[K-2]}{K-2} > \frac{b[K-1]}{K-1}$ . If the capacity of the knapsack is $\equiv 1 \pmod{K-1}$ , then we must use at least $K-2$ copies of $b[K-2]$ to fully fill the knapsack, occupying a total space of $(K-2)^2 = K^2-\mathcal{O}(K)$ .

Thus we can do the dynamic programming up to $K^2$ and do some math to simulate repeatedly using item $j$ after that.

Time Complexity: $\mathcal{O}(K^3)$

Subtask 5

Hint

Recall the subtask 2 solution. Can we apply a similar doubling idea?

Hint 2

It suffices to know $dp\left[\left\lceil \frac{i-K}{2} \right\rceil \dots \left\lfloor \frac{i+K}{2} \right\rfloor\right]$ to calculate $dp[i]$ .

Solution

Note that it suffices to know $dp\left[\left\lceil \frac{i-K}{2} \right\rceil \dots \left\lfloor \frac{i+K}{2} \right\rfloor\right]$ to calculate $dp[i]$ , as we can partition the items used in forming $dp[i]$ into two subsets with sums between $\left\lceil \frac{i-K}{2} \right\rceil$ and $\left\lfloor \frac{i+K}{2} \right\rfloor$ . If we were to compute $dp[N]$ recursively in this way, then the only states we visit are between $\frac{N}{2^k}-K$ and $\frac{N}{2^k}+K$ for some $k \ge 0$ , and it follows that there are at most $\mathcal{O}(K \log N)$ states, with $\mathcal{O}(K)$ transitions each. Implementing this idea iteratively (in descending order of $k$ ) gives an $\mathcal{O}(K^2 \log N)$ solution.

A similar idea with a more similar flavour as the $\times 2+1$ trick is described here (Knapsack Speedup 3).

Time Complexity: $\mathcal{O}(K^2 \log N)$

Alternative Solution

Trying to extend the idea from subtask 4, it is natural to wonder: Does it suffice to only consider the top few $j$ with the smallest $\frac{b[j]}{j}$ ? It turns out that if we only consider the $\left\lceil \frac{2K^2}{i} \right\rceil+1$ best $j$ when calculating $dp[i]$ , then this solution is provably correct.

Proof

Suppose this is false for the first time at $i$ . Let $m = \left\lceil \frac{2K^2}{i} \right\rceil+1$ , let $j_1, j_2, \dots, j_m$ be the $m$ $j$ with the smallest $\frac{b[j]}{j}$ , and let $u_1, u_2, \dots, u_k$ be the items we used to form $dp[i]$ . The main idea is that we will find a nonempty subset of $u_1, u_2, \dots, u_k$ that can be represented as the sum of a linear combination of $j_1, j_2, \dots, j_m$ , each coefficient being positive. Towards this, note that:

A theorem of Erdos and Graham (1972) on Frobenius numbers states that for any sequence $p_1 < p_2 < \dots < p_n$ ~p_1 < p_2 < \dots < p_n~ with $\gcd(p_1, p_2, \dots, p_n) = 1$ ~\gcd(p_1, p_2, \dots, p_n) = 1~, every number larger than $\frac{2p_1p_n}{n}-p_1$ ~\frac{2p_1p_n}{n}-p_1~ can be represented as the sum of a linear combination of $p_1, p_2, \dots, p_n$ ~p_1, p_2, \dots, p_n~ with all positive coefficients. We can extend this theorem to the case when $g = \gcd(p_1, p_2, \dots, p_n) \ne 1$ ~g = \gcd(p_1, p_2, \dots, p_n) \ne 1~ by applying the theorem to $\frac{p_1}{g}, \frac{p_2}{g}, \dots, \frac{p_n}{g}$ , giving the result that every number larger than $\frac{2p_1p_n}{gn}-p_1$ ~\frac{2p_1p_n}{gn}-p_1~ can be represented in this way.
To apply this theorem, we need to first bound $g = \gcd(j_1, j_2, \dots, j_m)$ ~g = \gcd(j_1, j_2, \dots, j_m)~. Sort $j_1, j_2, \dots, j_m$ ~j_1, j_2, \dots, j_m~ in increasing order of $j$ ~j~. As $j_1, j_2, \dots, j_m$ ~j_1, j_2, \dots, j_m~ are all between $1$ ~1~ and $K-1$ ~K-1~, some difference of adjacent terms $d = j_{x+1}-j_x$ ~d = j_{x+1}-j_x~ is at most $\frac{K-2}{m-1}$ ~\frac{K-2}{m-1}~. It follows that $g \le \frac{K-2}{m-1}$ ~g \le \frac{K-2}{m-1}~.
Now we need to find a subset of $u_1, u_2, \dots, u_k$ ~u_1, u_2, \dots, u_k~ with a large sum that is a multiple of $g$ ~g~. Recall the lemma that every set of $k \ge g$ ~k \ge g~ numbers has a subset that sums to a multiple of $g$ ~g~. We can repeatedly use this lemma to extract a subset of unused elements of $u_1, u_2, \dots, u_k$ ~u_1, u_2, \dots, u_k~, summing to a multiple of $g$ ~g~, as long as the sum of the remaining unused elements is at least $(K-1)(g-1)+1$ ~(K-1)(g-1)+1~. This way, we can find a subset $S$ ~S~ of $u_1, u_2, \dots, u_k$ ~u_1, u_2, \dots, u_k~ that sums to a multiple of $g$ ~g~ and is at least $i-(K-1)(g-1)$ ~i-(K-1)(g-1)~.
To apply the theorem to $j_1 < j_2 < \dots < j_m$ ~j_1 < j_2 < \dots < j_m~ and the sum of elements in $S$ ~S~, we need to check that $\displaystyle i-(K-1)(g-1) > \frac{2j_1j_m}{gm}-j_1$ Which follows from $\displaystyle g \le \frac{K-2}{m-1} < \frac{K}{\frac{2K^2}{i}} = \frac{i}{2K} < \frac{i}{K-1} \implies \frac{i}{g} > K-1$ $\displaystyle \implies \frac{i(g-1)}{g} \ge (K-1)(g-1) \implies i - \frac{i}{g} \ge (K-1)(g-1)$ $\displaystyle \implies i-(K-1)(g-1) \ge \frac{i}{g} > \frac{2K^2}{gm} > \frac{2j_1j_m}{gm}-j_1$ as needed.

Now replacing this subset with copies of items $j_1, j_2, \dots, j_m$ as needed gives the contradiction.

Time Complexity: $\mathcal{O}(K^2 \log K)$

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