Wise Men (Hard Version)

有 $n$ 个人。给出 $n$ 个人的「认识情况」(双向且保证合法)。 定义 `bool` 值函数 $K(i,j)$ 表示两个人是否认识。考虑一个 $n$ 个人编号的排列 $p$ ，定义其生成的 $01$ 串 $\{s_{n-1}\}$ 为 $\forall i\in[1,n-1]\cap\mathbb{Z_+},s_i=K(p_{i},p_{i+1})$。 统计对于 $2^{n-1}$ 种不同的 $01$ 串，有多少种排列可以生成它。 $2\leq n\leq 18$ 。

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.

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.

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 $ .

3
011
101
110

0 0 0 6

4
0101
1000
0001
1010

2 2 6 2 2 6 2 2

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.