PROBLEM LINK:

Practice
Contest

Author:Yuri Shilyaev
Tester:Hasan Jaddouh
Editorialist:Yury Shilyaev

DIFFICULTY:

Easy-medium.

PREREQUISITES:

Xor operation, dfs.

PROBLEM:

You are given a tree with $N$ vertices and $N - 1$ edge. Each vertex $i$ has a value $a_i$ associated with it. You have to count the number of unordered pairs $(u, v)$, that the parity of the number of one-bits in $a_u \oplus a_v$ is not equal to the parity of the shortest path between $u$ and $v$.

QUICK EXPLANATION:

Let's use the fact that parity of the number of one-bits $a_u \oplus a_v$ will not change if we replace $a_u$ with parity of one bits in $a_u$.

EXPLANATION:

Let $|x|$ be the number of one-bits in binary representation of $x$, $dist(u, v)$ be the length of the shortest path between $u$ and $v$, and $x\ \%\ 2$ be the remainder of the division $x$ by two.

It is easy to prove the fact, that $|a_u \oplus a_v|\ \%\ 2 = (|a_u|\ \%\ 2) \oplus (|a_v|\ \%\ 2)$. To use it, let's transform $b_i = |a_i|\ \%\ 2$.

Now the task is the following: count the number of pairs $(u, v)$, that $b_u \oplus b_v \neq dist(u, v)\ \%\ 2$. If we had rooted our tree, calculated for each $v$ it's $dist(root, v)$ (the number of edges in the path from the root), then:

$dist(u, v)\ \%\ 2 = (dist(root, u)\ \%\ 2) \oplus (dist(root, v)\ \%\ 2)$.

That's why we apply transform $d_i = dist(root, i) \ \%\ 2$ for array $d$ and reduce a problem to an easy one: count the number of pairs $(u, v)$, such that $b_u \oplus b_v \neq d_u \oplus d_v$.

By transfering everything to the left we achieve:

$b_u \oplus b_v \oplus d_u \oplus d_v = 1$

Easy problem, right? Let's just calc $cnt[bvalue][dvalue]$ $-$ number of vertices with $b_u = bvalue$ and $d_u = dvalue$, and iterate all possible pairs of $bvalue$ and $dvalue$ that give xor 1, and add $cnt[bvalue_u][dvalue_u] \cdot cnt[bvalue_v][dvalue_v]$ to the answer.

AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution can be found here.
Tester's solution can be found here.