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No.2911 Axioms of Topology

レベル : / AI推定難易度 : 650 / 実行時間制限 : 1ケース 2.000秒 / メモリ制限 : 512 MB / 標準ジャッジ問題
タグ : (AC するまで非表示) / 解いたユーザー数 39
作問者 : 👑 p-adic / テスター : hiro1729
お気に入りにしたユーザー ProblemId : 9308 / 自分の提出
問題文最終更新日: 2024-10-04 22:04:22
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yukicoder contest 449 (順位表) の他の問題:

Problem Statement

A string consisting only of the digits $0$ and $1$ is called a bit string.

The input gives $2$ positive integers $N, M$ and a set $S$ with $M$ elements, consisting only of bit strings of length $N$.

 

$S$ is said to be a good set of bit strings if it satisfies all of the following $3$ conditions:

  1. The string of length $N$ consisting only of the digit $1$ belongs to $S$.
  2. For any elements $s_0, s_1$ of $S$, the bitwise AND of $s_0$ and $s_1$ belongs to $S$.
  3. For any subset $S'$ of $S$, the bitwise OR of all elements of $S'$ belongs to $S$.

Here, in condition 3, the bitwise OR of all elements of $S'$ means the unique string $s$ of length $N$ consisting only of the digits $0$ and $1$ such that the following $2$ conditions hold for any positive integer $n$ not greater than $N$:

  • If the $n$-th character from the right of $s$ is $1$, then there exists an element $s'$ of $S'$ whose $n$-th character from the right is $1$.
  • If the $n$-th character from the right of $s$ is $0$, then there does not exist an element $s'$ of $S'$ whose $n$-th character from the right is $1$.

 

Determine whether $S$ is a good set of bit strings.

Background

Let $I_N$ be the set of all positive integers not greater than $N$, for each $s \in S$ let $U_s$ be the set of all $i \in I_N$ satisfying the condition "the $i$-th character from the right of $s$ is $1$", and let $\mathcal{O}_S$ be the family of sets $\{U_s \mid s \in S\}$. In this way, $S$ can be translated into the family $\mathcal{O}_S$ of subsets of $I_N$.

When geometry is formulated in modern set theory, the concept of a topological space is often used. The pair $(I_N,\mathcal{O}_S)$ is a topological space if and only if it satisfies all of the following $3$ conditions:

  1. $I_N$ belongs to $\mathcal{O}_S$.
  2. For any elements $U_0, U_1$ of $\mathcal{O}_S$, $U_0 \cap U_1$ belongs to $\mathcal{O}_S$.
  3. For any subset $\mathcal{U}$ of $\mathcal{O}_S$, $\bigcup_{U \in \mathcal{U}} U$ belongs to $\mathcal{O}_S$.

Therefore, by definition, this condition is equivalent to $S$ being a good set of bit strings. In other words, this problem is equivalent to determining whether $(I_N,\mathcal{O}_S)$ is a topological space.

Input

Number the elements of $S$ by the positive integers $m$ not greater than $M$, and denote them by $s_m$.

Then, the input is given from Standard Input in $1 + M$ lines in the following format:

  • On line $1$, $N, M$ are given, separated by a half-width space.
  • For each positive integer $m$ not greater than $M$, $s_m$ is given on line $1 + m$.
$N$ $M$
$s_1$
$\vdots$
$s_M$

Constraints

The input satisfies the following constraints:

  • $N$ is an integer satisfying $1 \leq N \leq 16$
  • $M$ is an integer satisfying $1 \leq M \leq \min \{2^N,2^{12}\}$
  • For any positive integer $m$ not greater than $M$, $s_m$ is a string of length $N$ consisting only of the digits $0$ and $1$

Output

Print Yes if $S$ is a good set of bit strings, and No otherwise.

Print a newline at the end.

Sample

Sample 1
Input
1 1
0
Output
No

$S = \{0\}$ does not satisfy condition 1. Indeed, $S$ has no element other than $0$, and the string $0$ is not a string consisting only of the digit $1$.

Also, $(I_N,\mathcal{O}_S) = (\{1\},\{\{\}\})$. Since $I_N \notin \mathcal{O}_S$, $(I_N,\mathcal{O}_S)$ is not a topological space.

Sample 2
Input
1 1
1
Output
No

$S = \{1\}$ does not satisfy condition 3. Indeed, if the subset $S'$ of $S$ is taken to be the empty set, its bitwise OR is $0$, but $0$ does not belong to $S$.

Also, $(I_N,\mathcal{O}_S) = (\{1\},\{\{1\}\})$. Letting $\mathcal{U} := \{\}$, we have $\mathcal{U} \subset \mathcal{O}_S$, but $\bigcup_{U \in \{\}} U = \{\} \notin \mathcal{O}_S$, so $(I_N,\mathcal{O}_S)$ is not a topological space.

Sample 3
Input
1 2
0
1
Output
Yes

$S = \{0,1\}$ satisfies all of conditions 1,2,3, so it is a good set of bit strings.

Also, $(I_N,\mathcal{O}_S) = (\{1\},\{\{\},\{1\}\})$. For those who know topological spaces, $(I_N,\mathcal{O}_S)$ is both a discrete space and an indiscrete space. For those who know algebraic geometry, $(I_N,\mathcal{O}_S)$ is a topological space homeomorphic to the affine scheme $\textrm{Spec}(\mathbb{Q}_p)$ associated with the field $\mathbb{Q}_p$ of $p$-adic numbers.

Sample 4
Input
3 4
000
011
110
111
Output
No

$S = \{000,011,110,111\}$ does not satisfy condition 2. Indeed, $011$ and $110$ are elements of $S$, but their bitwise AND $010$ does not belong to $S$.

Also, $(I_N,\mathcal{O}_S) = (\{1,2,3\},\{\{\},\{2,3\},\{1,2\},\{1,2,3\}\})$. $\{2,3\}$ and $\{1,2\}$ are elements of $\mathcal{O}_S$, but $\{2,3\} \cap \{1,2\} = \{1\} \notin \mathcal{O}_S$, so $(I_N,\mathcal{O}_S)$ is not a topological space.

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