We were waiting several weeks tp setting this contest and hope the problem was good enougho setting this contest and hope the problem was good enough.↵
### Events↵
↵
There was a difficulty with [problem:805E]/[problem:804C]. A little bug in the checker fortunately yields accepting an incorrect solution of only one person during the contest. I should apologize all of you because of this.↵
↵
For this sentence, `Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` You can find the meaning of "connected subgraph" with [connected](http://mathworld.wolfram.com/ConnectedGraph.html) and [subgraph](http://www.edmath.org/MATtours/discrete/concepts/csubgr.html), thus it can be empty as it is more logical. How ever, I should apologize all of the participants because of weak sample tests in the statement.↵
↵
I will write the full editorial in the few next days, now some hints and short solutions exist here.↵
↵
[problem:805A]↵
↵
<spoiler summary="hint1">↵
Almost half of numbers are divisible by two.↵
</spoiler>↵
↵
[problem:805B]↵
↵
<spoiler summary="hint1">↵
For the string shouldn't be any $i \leq n-2$, $s_i = s_{i+2}$. Make some examples of little sizes.↵
</spoiler>↵
↵
[problem:805C] / [problem:804A]↵
↵
<spoiler summary="hint1">↵
Find minimum weighted edges.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
There is some edges with weight 0, try to connect them in a carefully way.↵
</spoiler>↵
↵
[problem:805D] / [problem:804B]↵
↵
<spoiler summary="hint1">↵
The last situation is some $'a'$ characters after some $'b'$ ones.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The last situation is unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
The number of steps is also unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
Each $'b'$ character makes a number of $'b'$ characters in the last situation according to the number of $'a'$ characters before it. ↵
</spoiler>↵
↵
[problem:805E] / [problem:804C]↵
↵
<spoiler summary="hint1">↵
We want to color the ice creams in a way that all of ice creams in a vertex would be differentandwith this condition that if two vertices &u&$u$ and &v&$v$ have ice cream $i$, then all of vertices in the unique path of $u$ and $v$ would have it.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's have a trivial upper-bound for the answer.↵
↵
<spoiler summary="answer">↵
The trivial upper-bound should be maximum size of vertices' sets.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
You can prove that by induction on leaves of the tree. And code it how you proved it. ↵
</spoiler>↵
↵
There were some opinion about this sentence of the problem: ↵
↵
`Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` ↵
↵
A subgraph $S$ of a graph $G$ is a graph whose set of vertices and set of edges are all subsets of $G$. And according to this definition, the subgraph can also be empty as a subset of vertices and edges.↵
↵
A graph which is connected in the sense of a topological space, i.e., there is a path from any point to any other point in the graph. A graph that is not connected is said to be disconnected. This definition means that the null graph and singleton graph are considered connected, while empty graphs on $2 \leq n$ nodes are disconnected. ↵
↵
All of these definitions were from valid articles and it seems more logical. How ever, I should apologize all of the participants because of bad sample tests when some of articles and books didn't accept my opinion.↵
↵
[problem:805F] / [problem:804D]↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. consider the trivial $\mathcal{O}{(n^2)}$ solution and try to improve it.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually,">↵
seems SQRT Decomposition.↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
[problem:805F] / [problem:804D]↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. ↵
↵
<spoiler summary="Trivail solution">↵
$d_t$ is diameter of $t$-th tree and $f_i$ is the maximum path starting from vertex $i$, for all valid $i$ and $j$, assume that the edge is between them and find the diameter with $\max(f_i + f_j + 1, \max(d_1, d_2))$.↵
</spoiler>↵
↵
↵
consider this trivial $\mathcal{O}{(n^2)}$ solution and try to improve it to $\mathcal{O}{(n\cdot \log{(n)})}$.↵
</spoiler>↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually">↵
seems SQRT. ↵
↵
<spoiler summary="How?">↵
For every new query, try to do it with the complexity of $\mathcal{O}{( \min(sz_u, sz_v) \cdot log{(\max{( sz_{u} , sz_{v} )}}) )}$ , which $sz_i$ means size of the components of $i$-th vertex. Prove its complexity is $\mathcal{O}{(n \cdot \sqrt{(n)} \cdot \log{(n)})}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
You can prove, we need even number of swaps to reach the same permutation. So ${n \choose 2} \mod 2 = 0$ and the answer for $4 \cdot k + 3$ and $4 \cdot k+4$ is -1.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The solution almost is constructive.↵
↵
<spoiler summary="Try to solve it for n = 4">↵
And then solve $n=8$ by $n=4$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
Let's partition $n=4\cdot k$ numbers to classes, each class contains a $4$ consecutive numbers. And when $n=4\cdot k+1$ there is a class with $5$ consecutive numbers.↵
</spoiler>↵
↵
[problem:804F]↵
↵
<spoiler summary="hint1">↵
The problem consists two parts, the first part is to find $mn_i$ and $mx_i$, the minimum and the maximum score $i$-th vertex may get after the score distribution. The second part is to finding the answer and all your need to solve this part is those two arrays. It is trivial $mn_i$ is number of real bullion of golds in $i$-th vertex.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's find the scores for two vertices $u$ and $v$, $u$ have a directed edge to $v$.↵
↵
<spoiler summary="u -> v">↵
$$g = \gcd(s_u, s_v)$$↵
If $i$-th element from $u$-th vertex was painted and $j$-th element from $v$-th vertex was unpainted such that $i \mod g \equiv j \mod g$ then $j$-th element from $v$-th vertex will be painted quickly. For two vertex, we can do it in complexity $\mathcal{O}{(s_u+s_v)}$.↵
</spoiler>↵
↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
If we have this directed walk ${w_1, w_2, \... , w_k}$, you should affect $v_1$ on $v_k$ how I said in the previous hint with $g = \gcd( s_{w_1}, s_{w_2}, ... , s_{w_k} )$.↵
↵
<spoiler summary="What about strogly connected components.">↵
$$g = \gcd( s_{1} , s_{2} , ... , s_{n} )$$↵
In this components if $i$-th element from $u$-th vertex was painted then for all of $j$ such that $i \mod g \equiv j \mod g$, $j$-th element from all vertices will be painted quickly. For this components, we can do it with complexity $\mathcal{O}{({\sum_{i=1}^{n} s_i})}$. In the main problem, we can suppose all of a strongly conected component a vertex with size $g$ in this way and we can calculate $mx_i$ with answer of its component's vertex multiplied by $\frac{s_i}{g}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
SCC of the tournament yields a Hamiltonian path of strongly connected components( if we consider each component a vertex ). The problem decreased to solve this path. we can solve the path with the complexity of sum of vertices' sizes.↵
</spoiler>↵
↵
↵
<spoiler summary="hint5">↵
Now, for every vertex we have minimum and maximum score it may get. For B known vertices, to check if they can be the wanted set, we can assume their score be maximum possible and the other scores be minimum possible. If they were top, then they can be a sample of the wanted set.↵
</spoiler>↵
↵
↵
<spoiler summary="hint6">↵
Consider a sorted sequence of combination of two array $mn$ and $mx$. ↵
↵
For every vertex, consider the number of ways it(with its maximum score) can be the minimum score of the set of B vertices.↵
</spoiler>↵
↵
### Events↵
↵
There was a difficulty with [problem:805E]/[problem:804C]. A little bug in the checker fortunately yields accepting an incorrect solution of only one person during the contest. I should apologize all of you because of this.↵
↵
For this sentence, `Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` You can find the meaning of "connected subgraph" with [connected](http://mathworld.wolfram.com/ConnectedGraph.html) and [subgraph](http://www.edmath.org/MATtours/discrete/concepts/csubgr.html), thus it can be empty as it is more logical. How ever, I should apologize all of the participants because of weak sample tests in the statement.↵
↵
I will write the full editorial in the few next days, now some hints and short solutions exist here.↵
↵
[problem:805A]↵
↵
<spoiler summary="hint1">↵
Almost half of numbers are divisible by two.↵
</spoiler>↵
↵
[problem:805B]↵
↵
<spoiler summary="hint1">↵
For the string shouldn't be any $i \leq n-2$, $s_i = s_{i+2}$. Make some examples of little sizes.↵
</spoiler>↵
↵
[problem:805C] / [problem:804A]↵
↵
<spoiler summary="hint1">↵
Find minimum weighted edges.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
There is some edges with weight 0, try to connect them in a carefully way.↵
</spoiler>↵
↵
[problem:805D] / [problem:804B]↵
↵
<spoiler summary="hint1">↵
The last situation is some $'a'$ characters after some $'b'$ ones.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The last situation is unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
The number of steps is also unique.↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
Each $'b'$ character makes a number of $'b'$ characters in the last situation according to the number of $'a'$ characters before it. ↵
</spoiler>↵
↵
[problem:805E] / [problem:804C]↵
↵
<spoiler summary="hint1">↵
We want to color the ice creams in a way that all of ice creams in a vertex would be different
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's have a trivial upper-bound for the answer.↵
↵
<spoiler summary="answer">↵
The trivial upper-bound should be maximum size of vertices' sets.↵
</spoiler>↵
↵
</spoiler>↵
↵
You can prove that by induction on leaves of the tree. And code it how you proved it. ↵
</spoiler>↵
↵
↵
`Vertices which have the i-th (1 ≤ i ≤ m) type of ice cream form a connected subgraph.` ↵
↵
A subgraph $S$ of a graph $G$ is a graph whose set of vertices and set of edges are all subsets of $G$. And according to this definition, the subgraph can also be empty as a subset of vertices and edges.↵
↵
A graph which is connected in the sense of a topological space, i.e., there is a path from any point to any other point in the graph. A graph that is not connected is said to be disconnected. This definition means that the null graph and singleton graph are considered connected, while empty graphs on $2 \leq n$ nodes are disconnected. ↵
↵
All of these definitions were from valid articles and it seems more logical. How ever, I should apologize all of the participants because of bad sample tests when some of articles and books didn't accept my opinion.↵
↵
[problem:805F] / [problem:804D]↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. consider the trivial $\mathcal{O}{(n^2)}$ solution and try to improve it.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually,">↵
seems SQRT Decomposition.↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
↵
<spoiler summary="hint1">↵
Let's solve the problem for just two tree. ↵
↵
<spoiler summary="Trivail solution">↵
$d_t$ is diameter of $t$-th tree and $f_i$ is the maximum path starting from vertex $i$, for all valid $i$ and $j$, assume that the edge is between them and find the diameter with $\max(f_i + f_j + 1, \max(d_1, d_2))$.↵
</spoiler>↵
↵
↵
consider this trivial $\mathcal{O}{(n^2)}$ solution and try to improve it to $\mathcal{O}{(n\cdot \log{(n)})}$.↵
</spoiler>↵
↵
<spoiler summary="hint2">↵
$$\sum_{i=1}^{n} sz_i = n$$↵
↵
<spoiler summary="Actually">↵
seems SQRT. ↵
↵
<spoiler summary="How?">↵
For every new query, try to do it with the complexity of $\mathcal{O}{( \min(sz_u, sz_v) \cdot log{(\max{( sz_{u} , sz_{v} )}}) )}$ , which $sz_i$ means size of the components of $i$-th vertex. Prove its complexity is $\mathcal{O}{(n \cdot \sqrt{(n)} \cdot \log{(n)})}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
</spoiler>↵
↵
[problem:804E]↵
↵
<spoiler summary="hint1">↵
You can prove, we need even number of swaps to reach the same permutation. So ${n \choose 2} \mod 2 = 0$ and the answer for $4 \cdot k + 3$ and $4 \cdot k+4$ is -1.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
The solution almost is constructive.↵
↵
<spoiler summary="Try to solve it for n = 4">↵
And then solve $n=8$ by $n=4$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
Let's partition $n=4\cdot k$ numbers to classes, each class contains a $4$ consecutive numbers. And when $n=4\cdot k+1$ there is a class with $5$ consecutive numbers.↵
</spoiler>↵
↵
[problem:804F]↵
↵
<spoiler summary="hint1">↵
The problem consists two parts, the first part is to find $mn_i$ and $mx_i$, the minimum and the maximum score $i$-th vertex may get after the score distribution. The second part is to finding the answer and all your need to solve this part is those two arrays. It is trivial $mn_i$ is number of real bullion of golds in $i$-th vertex.↵
</spoiler>↵
↵
↵
<spoiler summary="hint2">↵
Let's find the scores for two vertices $u$ and $v$, $u$ have a directed edge to $v$.↵
↵
<spoiler summary="u -> v">↵
$$g = \gcd(s_u, s_v)$$↵
If $i$-th element from $u$-th vertex was painted and $j$-th element from $v$-th vertex was unpainted such that $i \mod g \equiv j \mod g$ then $j$-th element from $v$-th vertex will be painted quickly. For two vertex, we can do it in complexity $\mathcal{O}{(s_u+s_v)}$.↵
</spoiler>↵
↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint3">↵
If we have this directed walk ${w_1, w_2, \... , w_k}$, you should affect $v_1$ on $v_k$ how I said in the previous hint with $g = \gcd( s_{w_1}, s_{w_2}, ... , s_{w_k} )$.↵
↵
<spoiler summary="What about strogly connected components.">↵
$$g = \gcd( s_{1} , s_{2} , ... , s_{n} )$$↵
In this components if $i$-th element from $u$-th vertex was painted then for all of $j$ such that $i \mod g \equiv j \mod g$, $j$-th element from all vertices will be painted quickly. For this components, we can do it with complexity $\mathcal{O}{({\sum_{i=1}^{n} s_i})}$. In the main problem, we can suppose all of a strongly conected component a vertex with size $g$ in this way and we can calculate $mx_i$ with answer of its component's vertex multiplied by $\frac{s_i}{g}$.↵
</spoiler>↵
↵
</spoiler>↵
↵
↵
<spoiler summary="hint4">↵
SCC of the tournament yields a Hamiltonian path of strongly connected components( if we consider each component a vertex ). The problem decreased to solve this path. we can solve the path with the complexity of sum of vertices' sizes.↵
</spoiler>↵
↵
↵
<spoiler summary="hint5">↵
Now, for every vertex we have minimum and maximum score it may get. For B known vertices, to check if they can be the wanted set, we can assume their score be maximum possible and the other scores be minimum possible. If they were top, then they can be a sample of the wanted set.↵
</spoiler>↵
↵
↵
<spoiler summary="hint6">↵
Consider a sorted sequence of combination of two array $mn$ and $mx$. ↵
↵
For every vertex, consider the number of ways it(with its maximum score) can be the minimum score of the set of B vertices.↵
</spoiler>↵
↵
