Wise Men (Hard Version)
Contest: Codeforces Global Round 7
Difficulty rating: 3200
Tags: bitmasks, dp, math

This is the hard version of the problem. The difference is constraints on the number of wise men and the time limit. You can make hacks only if all versions of this task are solved.

$$$n$$$ wise men live in a beautiful city. Some of them know each other.

For each of the $$$n!$$$ possible permutations $$$p_1, p_2, \ldots, p_n$$$ of the wise men, let's generate a binary string of length $$$n-1$$$: for each $$$1 \leq i < n$$$ set $$$s_i=1$$$ if $$$p_i$$$ and $$$p_{i+1}$$$ know each other, and $$$s_i=0$$$ otherwise.

For all possible $$$2^{n-1}$$$ binary strings, find the number of permutations that produce this binary string.

Input format:
The first line of input contains one integer $$$n$$$ ($$$2 \leq n \leq 18)$$$  — the number of wise men in the city.

The next $$$n$$$ lines contain a binary string of length $$$n$$$ each, such that the $$$j$$$-th character of the $$$i$$$-th string is equal to '1' if wise man $$$i$$$ knows wise man $$$j$$$, and equals '0' otherwise.

It is guaranteed that if the $$$i$$$-th man knows the $$$j$$$-th man, then the $$$j$$$-th man knows $$$i$$$-th man and no man knows himself.

Output format:
Print $$$2^{n-1}$$$ space-separated integers. For each $$$0 \leq x < 2^{n-1}$$$:

- Let's consider a string $$$s$$$ of length $$$n-1$$$, such that $$$s_i = \lfloor \frac{x}{2^{i-1}} \rfloor \bmod 2$$$ for all $$$1 \leq i \leq n - 1$$$.
- The $$$(x+1)$$$-th number should be equal to the required answer for $$$s$$$.

Constraints:
Time limit: 4.0 s
Memory limit: 512.0 MB

Note:
In the first test, each wise man knows each other, so every permutation will produce the string $$$11$$$.

In the second test:

- If $$$p = \{1, 2, 3, 4\}$$$, the produced string is $$$101$$$, because wise men $$$1$$$ and $$$2$$$ know each other, $$$2$$$ and $$$3$$$ don't know each other, and $$$3$$$ and $$$4$$$ know each other;
- If $$$p = \{4, 1, 2, 3\}$$$, the produced string is $$$110$$$, because wise men $$$1$$$ and $$$4$$$ know each other, $$$1$$$ and $$$2$$$ know each other and $$$2$$$, and $$$3$$$ don't know each other;
- If $$$p = \{1, 3, 2, 4\}$$$, the produced string is $$$000$$$, because wise men $$$1$$$ and $$$3$$$ don't know each other, $$$3$$$ and $$$2$$$ don't know each other, and $$$2$$$ and $$$4$$$ don't know each other.

Example input:
3
011
101
110

Example output:
0 0 0 6

Example input:
4
0101
1000
0001
1010

Example output:
2 2 6 2 2 6 2 2

Editorial:
For each binary string $$$s$$$, let's calculate $$$f(s)$$$ – the number of permutations, such that, if $$$s_i=1$$$, then $$$p_i$$$ and $$$p_{i+1}$$$ know each other, otherwise, they may know or don't know each other.
To get real answers, we may use inclusion-exclusion, which may be optimized using a straightforward sum over subsets dp.
To calculate $$$f(s)$$$, we need to note that $$$f$$$ depends only on the multiset of lengths of blocks of $$$1$$$'s in it.
For example, $$$f(110111) = f(111011)$$$, because the multiset of lengths is $$$\{1, 3, 4\}$$$ (note that block of size $$$x$$$ of $$$1$$$'s corresponds to length $$$x+1$$$).
And note that there are exactly $$$P(n)$$$ (the number of partitions of $$$n$$$) possible multisets.
$$$P(18) = 385$$$
To process further, at first let's calculate $$$g_{len, mask}$$$ – the number of paths of length $$$len$$$, which pass only through the vertices from $$$mask$$$ (and only through them).
You can calculate it with a straightforward $$$dp_{mask, v}$$$ in $$$\mathcal{O}{(2^n \cdot n^2)}$$$.
Then, let's fix the multiset of lengths $$$a_1, a_2, \ldots, a_k$$$.
I claim that the $$$f(s)$$$ for this multiset is equal to $$$\sum{\prod{g_{a_i, m_i}}}$$$ over all masks $$$m_1, m_2, \ldots m_k$$$, such that the bitwise OR of $$$m_1, m_2, \ldots, m_k$$$ is equal to $$$2^n-1$$$ (note that we don't care about the number of bits like in a usual non-intersecting subsets convolution, because if some masks are intersecting, then their OR won't be equal to $$$2^n-1$$$ because $$$\sum{a_i} = n$$$).
You can calculate this sum by changing $$$g_{len}$$$ to the sum over subsets.
And then, for this partition, you can just calculate $$$d_{mask} = \prod{g_{a_i, mask}}$$$ in $$$\mathcal{O}{(k \cdot 2^n)}$$$, and you can restore the real value of $$$2^n-1$$$ by inclusion-exclusion in $$$\mathcal{O}{(2^n)}$$$.
If you will calculate this naively, you will get the $$$\mathcal{O}$$$((sum of sizes of all partitions) $$$\cdot 2^n)$$$ solution, which is enough to get AC.
But you can optimize this because you can maintain $$$d_{mask}$$$ during the brute force of all partitions. And in the tree of all partitions, there are $$$\mathcal{O}{(P(n))}$$$ intermediate vertices, so it will work in $$$\mathcal{O}{(P(n) \cdot 2^n)}$$$.
The total complexity is $$$\mathcal{O}{((P(n) + n^2) \cdot 2^n))}$$$.