A13292 | Wise Men (Easy Version)
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题目描述
This is the easy 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.
$n$ 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.
输入格式
The first line of input contains one integer $n$ ( $2 \leq n \leq 14)$ — 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.
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.
输出格式
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 $s$ .
- 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 $s$ .
输入输出样例
输入 #1
3 011 101 110
输出 #1
0 0 0 6
输入 #2
4 0101 1000 0001 1010
输出 #2
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.
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.
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评测结果：Accepted