#### A — Bath Temperature

**Problem author:** teochaban

**Solution**

**Editorial author:** tomasf

If any tap has the desired temperature, there is no need to open a second tap, so the answer is 1.

Since the resulting temperature is a weighted mean of the opened taps, all obtainable temperatures are between the minimum and maximum temperature available on the taps. If the desired temperature is outside that range, the answer is -1. Otherwise, we can use the minimum and maximum temperature to obtain the desired temperature, so the answer is 2.

**Overall Complexity:** *O*(*N*)

#### B — Nicoleta's Cleaning

**Problem author:** tomasf

**Solution**

**Editorial author:** tomasf

This problem can be solved with a sweep line approach.

First, compress all *y* coordinates, the resulting coordinates shouldn't be greater than 3·10^{5} since this is the maximum number of distinct *y* coordinates given in the input. Then sort all points and vertical lines of the rectangles (events) by the *x* coordinate and process them sequentially. Use a range-update/point-query data structure (like a Fenwick Tree or a Segment Tree) to keep how many rectangles are "active" for every *y* coordinate. For every vertical line event, indicating the start or the end of a rectangle, update the data structure on the related *y* coordinate range. For every point event, query the data structure on the related *y* coordinate and store the answer.

Special care must be taken with indexing since the problem asks for the number of rectangles in which the point is **strictly** inside. Also, if there are points and vertical lines on the same *x* coordinate, they must be processed in a specific order so that rectangles that have just started or finished are ignored. The order is: Rectangle end event, point event, rectangle start event.

**Overall Complexity:**

#### C — Leading the Scoreboard

**Problem author:** tomasf

**Solution**

**Editorial author:** tomasf

This problem can be solved by simulating the scoreboard. One of the easiest ways of implementing that is by processing the submissions sequentially while keeping track of what team is leading the scoreboard, how many problems they solved and their total time (penalty). Special care must be taken to handle the occasions in which multiple teams are leading the scoreboard.

**Overall Complexity:** *O*(*M*)

**Code(nonsequitur):** Link

#### D — Joaozao, The Digit Maker

**Problem author:** danft

**Solution**

**Editorial author:** danft

Let *A* be the sequence of digits ordered non-decreasingly. Also, let's assume that *A* is composed of distinct elements. Let and let *G* be the functional graph of *f*.

#### There is at least one cycle of size 2 in *G*.

- If there is a pair (
*d*_{i},*d*_{j}), such that,*d*_{i}+*d*_{j}<*B*. Then it is obvious that if*f*(*i*) =*k*, then*f*(*k*) =*i*, and*d*_{i}+*d*_{k}<*B*, as , for any*h*. - Otherwise,
*f*(2*n*) = 2*n*- 1 and*f*(2*n*- 1) = 2*n*(check it)

#### Solution

- Pick the size-2 cycle with greatest value
*z*= (*z*_{1}+*z*_{2})%*B*, - Add
*z*to the answer - Remove it from
*A*and rebuild*f*and*G*. - Do this until

#### Proof:

First thing to notice is that any other pairing will have value *w* ≤ *z*. This implies that the solution we are building is non-increasing. Suppose now that there is an optimal algorithm *O* that also builds its solution in non-increasing order. Let *O*_{k} be the set of available pairings *O* has available after step *k*, let also *B*_{k} be the set of pairings our algorithm (let's call it *B*) has after step *k*.

We are going to prove that *max*{*O*_{k}} = *max*{*B*_{k}}, for any *k* ≥ 0

Suppose that after step *k*, *max*{*O*_{k}} = *max*{*B*_{k}}. (base case *k* = 0 it is obviously true)

if *O* picks *w* < *z* it's not optimal, so it can't be, also by assumption *w* > *z* is not possible either.

if *O* picks *w* = *z* it is either the same cycle or a disjoint one (distinct elements assumption), as removing a cycle doesn't do anything to cycles already in *G*, *w* will be or has been picked by *B* and *z* will be or has been picked by *O*.

In either case, *max*{*O*_{k + 1}} = *max*{*B*_{k + 1}}.

This is sufficient to prove that our algorithm builds an answer as good as any other algorithm.

#### Observations:

Without assuming elements are distinct, *O* could remove a pairing which is not a cycle, the proof would need some small adjustments.

To simplify the implementation, we can remove cycles in any order, as removing one, does not affect others already in *G*. Also to simplify, let , if *f*(*i*) = *j* and *f*(*j*) = *i* and , then .

With both observations, we can implement a simpler solution which starts from *i* = 2*n*, and if , we skip it, otherwise we add *d*_{i} + *d*_{f(i)} to the answer. After that, we do the same for the elements we had skipped.

**Overall Complexity:** *O*(*nlogn*)

#### E — Double Fence

**Problem author:** tomasf

**Solution**

**Editorial author:** tomasf

Let *P*_{1} and *P*_{2} be respectively the set of points of the first polygon and the set of points of the second polygon. Let *ch*(*X*) be the set of points on the convex hull of the set of points *X*, including colinear points on the edges. A polygon is strictly inside the other if or .

**Overall Complexity:**

#### F — No Link, Cut Tree!

**Problem author:** teochaban

**Solution**

**Editorial author:** danft

First notice that because the binary tree is a complete binary tree, its height will be (less than 20).

Let's denote *d*_{u} as the depth of node *u* (its distance to the root), and *T*_{u} the set of nodes in *u*'s subtree.

Now let's define *S*[*u*][*d*] as the sum of shininess of nodes on depth *d*, which are in the subtree of *u*.

In other words,

Given a query *u*, the answer will be the depth *d*, such that:

As *S*[*u*][*d*] can be written in funciton of its children:

A simple dfs can then be used to compute it.

**Overall Complexity:** *O*(*Nlog*_{2}(*N*))

#### G — Hungry Canadian

**Problem author:** teochaban

**Solution**

**Editorial author:** danft

#### Dynamic Programing

The simplest solution for this problem uses dynamic programing on the recurrence:

*dp*[*i*][*j*] is the minimum cost of completing the string after having added *i* characters to it, *j* being the last one.

The answer will be:

*min*

_{0 ≤ j < 26}{

*dp*[1][

*j*]}

#### Matrix Exponentiation

Another solution uses matrix exponentiation. Let's define the operator × , which receives two squared matrices *A*, *B* and returns a squared matrix *C* with the same dimension *n*:

*C*=

*A*×

*B*

*C*

_{ij}=

*min*

_{1 ≤ k ≤ n}{

*A*

_{ik}+

*B*

_{kj}}

*P*^{k} (the matrix of costs to the power of *k*) = *P* × *P* × ... × *P*, will have the minimum cost to build a string of size *k*. More specifically *P*_{ij}^{k} is the minimum cost of building a string of size *k* starting with character *i* and ending with character *j*.

##### Proof

Let's prove it by induciton on *k*.

The base case *k* = 1 is correct by definition.

Now, assuming that *P*^{k} is the matrix of minimum costs of building a string of size *k*

(*P*^{k} × *P*)_{ij} = *min*_{1 ≤ k ≤ n}{*P*_{ik}^{k} + *P*_{kj}}

Because the string of minimum cost of length *k* + 1 is a string of minimum cost of length *k*, plus one character, *P*^{k + 1}_{ij} is the minimum cost of building a string of size *k* + 1 starting at *i* and ending at *j*. It can also be proven that × is associative, which allows us to use fast exponentiation algorithm.

**Overall Complexity:** *O*(*K*) or *O*(*log*(*K*))

#### H — Eating Pie

**Problem author:** bssanches

**Solution**

**Editorial author:** bssanches

This is a max flow problem. First, let's create a graph where every type of pie is a node. Then let's consider every pair of vertices *u* and *v*. They'll have an bi-directional edge if the pie with type *u* appears adjacent to the pie with type *v* in the array *P* and, the capacity of this edge, will be the sum of *G*_{i} in these positions. For example if *P* = {1, 3, 1, 2} and *G* = {1, 2, 4} then we will have and edge between 1 and 3 with capacity 3 and between 1 and 2 with capacity 4.

So for every pair of nodes *u* and *v* connected by an edge with capacity *C*, we know that the amount of candies someone is going to earn if they bought both *u* and *v* is *C*.

Now we are going to add two new nodes, one that is going to be the Joaozao (sink) and the other one the Nicoleta (source). We will assume now that every node that Joaozao reaches are going to be brought by him, the same goes for Nicoleta. Assuming this, we need to garantee that Joaozao reaches the nodes he is obliged to buy (the ones that does not appear in Nicoleta's list), the same for Nicoleta. So we will create an edge from Joaozao to every type of pie he is obliged to buy, this edge will have capacity ∞. We will do the same for Nicoleta.

With this graph now, every node has only 3 possible states:

The node cannot be reached by anyone

The node can only be reached by one person

The node can be reached by both Joaozao and Nicoleta.

For the first case, if the node cannot be reached, it means that the node can be brought by anyone and it is in an isolated component (because the edges are bi-directional). In that case anyone can buy the whole component (so the answer is only to sum all the edges in it).

For the second case, there's only one person that can reach the whole component. In that case the person that reaches should buy all the nodes. Which also means to sum all the edges in the answer.

For the third case, the component can be reached by both of them, so we need to separate it in two components. By doing this we will end up with the second case again. Note that by separating them in two components, we will be cutting some edges and they will not be part of our answer. So we want to cut such edges in a way that the sum of their capacities is minimized. This is a standard mincut/maxflow problem.

To sum it up, the answer is the sum of all edges given in the input (in other words, the sum of the array *G*) minus the maxflow.

**Overall Complexity:** Depends on the maxflow implementation, for the Dinic's algorithm it is:

*O*(*E*·*V*^{2})

Note that in the worst case, the complete graph will have around 40 nodes.

#### I — Matrix Sum

**Problem author:** bssanches

**Solution**

**Editorial author:** danft

Fist thing to notice is that the formula given in the problem statement gives us a clue about where to start. To make things a little bit cleaner let's define *R*, *M* and *S*:

Moving things around a bit we get:

*M*

_{ij}=

*C*

_{j}+

*R*

_{i}-

*A*

_{ij}

To find *C*_{j}, we sum the equation above over *i*.

The same thing can be done to find *R*_{i}. To get *S* we sum over the whole matrix *M*:

Now we can find *M*_{ij} using the first equation. Also, when dividing by *n* - 1 we assumed *n* > 1, so there's a corner case when *n* = 1. Also it is obvious that , if and only if .

**Overall Complexity:** *O*(*n*^{2})

#### J — Beautiful Triangles

**Problem author:** bimaoe

**Solution**

**Editorial author:** tomasf

Let *l*_{min} = *min*(*l*_{1}, *l*_{2}) and *l*_{max} = *max*(*l*_{1}, *l*_{2}). If the third side of the triangle (*l*_{3}) is shorter than *l*_{max}, the beauty *b* is given by *l*_{min} + *l*_{3} - *l*_{max} and can be increased by increasing *l*_{3}. Else, if the *l*_{3} is larger than *l*_{max}, the beauty *b* is given by *l*_{min} + *l*_{max} - *l*_{3} and can be increased by decreasing *l*_{3}. For that reason, the answer is given by *l*_{3} = *l*_{max} = *max*(*l*_{1}, *l*_{2}).

**Overall Complexity:** *O*(1)

#### K — Counting Good Teams

**Problem author:** tomasf

**Solution**

**Editorial author:** tomasf

Let's calculate the number of bad teams (teams that have at least one member that knows everything the other member knows) and subtract from the number of all possible teams. In other words, let's find out the number of distinct pairs *i*, *j* in which the bitmask of *x*_{i} is a submask of *x*_{j}.

To count the number of bad teams we can use a meet-in-the-middle approach. For every member *i*, let's split the bitmask *x*_{i} into *x*_{i}^{1} and *x*_{i}^{2}, the former containing the first bits of *x*_{i} and the latter containing the rest.

Let's define a concatenation operator that concatenates two bitmasks. For example: . Let's also define *S*_{↑}(*mask*) as the set of all bitmasks *x* of the same size of *mask* so that *mask* is a submask of *x*. Likewise, let's define *S*_{↓}(*mask*) as the set of all bitmasks *x* of the same size of *mask* so that *x* is submask of *mask*.

For pairs of distinct masks *x*_{i} and *x*_{j}, the intersection of with has exactly one element if *x*_{i} is submask of *x*_{j} and none otherwise.

Let's count the occurrence *cnt*_{k}^{↑} of each mask 0 ≤ *k* < 2^{M} when computing for every mask *x*_{i} given in the input, and do the same for , counting the occurrence *cnt*_{k}^{↓} of the mask *k*.

After handling with the pairs that have identical bitmasks, the number of bad teams is given by .

**Overall Complexity:** *O*(2^{M} + *N*·2^{M / 2})

**Code(tomasf):** Link, Link (alternative *O*(*N*·2^{M / 2}) meet-in-the-middle solution)