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internbootcamp/bootcamp/cquantifierquestion/cquantifierquestion.py
Executable file
335
internbootcamp/bootcamp/cquantifierquestion/cquantifierquestion.py
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"""#
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### 谜题描述
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Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let's focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal (∀) and existential (∃). You can read more about them here.
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The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
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* ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false.
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* ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true.
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The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
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* ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true.
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* ∃ x,x>x-1 is read as: there exists a real number x such that x is greater than x-1. This statement is true.
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Moreover, these quantifiers can be nested. For example:
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* ∀ x,∃ y,x<y is read as: for all real numbers x, there exists a real number y such that x is less than y. This statement is true since for every x, there exists y=x+1.
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* ∃ y,∀ x,x<y is read as: there exists a real number y such that for all real numbers x, x is less than y. This statement is false because it claims that there is a maximum real number: a number y larger than every x.
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Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
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There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
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where ∧ denotes logical AND. That is, f(x_1,…, x_n) is true if every inequality x_{j_i}<x_{k_i} holds. Otherwise, if at least one inequality does not hold, then f(x_1,…,x_n) is false.
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Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
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is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
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Note that the order the variables appear in the statement is fixed. For example, if f(x_1,x_2):=(x_1<x_2) then you are not allowed to make x_2 appear first and use the statement ∀ x_2,∃ x_1, x_1<x_2. If you assign Q_1=∃ and Q_2=∀, it will only be interpreted as ∃ x_1,∀ x_2,x_1<x_2.
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Input
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The first line contains two integers n and m (2≤ n≤ 2⋅ 10^5; 1≤ m≤ 2⋅ 10^5) — the number of variables and the number of inequalities in the formula, respectively.
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The next m lines describe the formula. The i-th of these lines contains two integers j_i,k_i (1≤ j_i,k_i≤ n, j_i≠ k_i).
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Output
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If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
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Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
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On the next line, output a string of length n, where the i-th character is \"A\" if Q_i should be a universal quantifier (∀), or \"E\" if Q_i should be an existential quantifier (∃). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.
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Examples
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Input
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2 1
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1 2
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Output
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1
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AE
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Input
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4 3
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1 2
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2 3
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3 1
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Output
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-1
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Input
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3 2
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1 3
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2 3
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Output
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2
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AAE
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Note
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For the first test, the statement ∀ x_1, ∃ x_2, x_1<x_2 is true. Answers of \"EA\" and \"AA\" give false statements. The answer \"EE\" gives a true statement, but the number of universal quantifiers in this string is less than in our answer.
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For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
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For the third test, the statement ∀ x_1, ∀ x_2, ∃ x_3, (x_1<x_3)∧ (x_2<x_3) is true: We can set x_3=max\\{x_1,x_2\}+1.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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import sys
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range = xrange
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input = raw_input
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def toposort(graph):
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n = len(graph)
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res = []
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found = [0] * n
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for i in range(n):
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if found[i]:
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continue
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stack = [i]
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while stack:
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node = stack.pop()
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if node < 0:
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res.append(~node)
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elif not found[node]:
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found[node] = 1
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stack.append(~node)
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for nei in graph[node]:
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if not found[nei]:
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stack.append(nei)
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# Check for cycle
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for node in res:
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if any(found[nei] for nei in graph[node]):
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return None
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found[node] = 0
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return res[::-1]
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inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
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n = inp[ii]; ii += 1
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m = inp[ii]; ii += 1
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coupl1 = [[] for _ in range(n)]
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coupl2 = [[] for _ in range(n)]
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for _ in range(m):
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u = inp[ii] - 1; ii += 1
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v = inp[ii] - 1; ii += 1
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coupl1[u].append(v)
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coupl2[v].append(u)
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order = toposort(coupl1)
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if order is None:
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print -1
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sys.exit()
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seen1 = list(range(n))
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seen2 = list(range(n))
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for node in order:
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for nei in coupl1[node]:
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seen1[nei] = min(seen1[node], seen1[nei])
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for node in reversed(order):
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for nei in coupl2[node]:
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seen2[nei] = min(seen2[node], seen2[nei])
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seen = [+(seen1[node] == seen2[node] == node) for node in range(n)]
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print sum(seen)
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print ''.join('A' if c else 'E' for c in seen)
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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import random
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import re
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from collections import deque
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class Cquantifierquestionbootcamp(Basebootcamp):
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def __init__(self, n_min=2, n_max=5, m_min=1, m_max=5):
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self.n_min = n_min
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self.n_max = n_max
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self.m_min = m_min
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self.m_max = m_max
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def case_generator(self):
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# 生成DAG的边
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n = random.randint(self.n_min, self.n_max)
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# 生成拓扑序
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top_order = list(range(1, n+1))
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random.shuffle(top_order)
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# 生成所有可能的边（仅从前到后）
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possible_edges = []
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for i in range(len(top_order)):
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for j in range(i+1, len(top_order)):
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possible_edges.append((top_order[i], top_order[j]))
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# 确定边的数量
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max_valid_edges = len(possible_edges)
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m = random.randint(max(self.m_min, 1), min(self.m_max, max_valid_edges))
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edges = random.sample(possible_edges, m) if possible_edges else []
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return {
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'n': n,
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'm': len(edges),
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'edges': edges
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}
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@staticmethod
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def prompt_func(question_case):
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edges = question_case['edges']
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edges_str = '\n'.join(f"{j} {k}" for j, k in edges)
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return f"""请为以下逻辑问题分配量词（A/E）使表达式为真且全称最多：
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变量数：{question_case['n']}，不等式数：{question_case['m']}
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不等式列表：
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{edges_str}
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答案格式：[answer]...[/answer]，若无解输出-1"""
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@staticmethod
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def extract_output(output):
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# 提取最后一个答案块
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answer_blocks = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not answer_blocks:
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return None
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content = answer_blocks[-1].strip()
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lines = [line.strip() for line in content.split('\n') if line.strip()]
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if not lines:
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return None
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# 处理-1的情况
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if lines[0] == '-1':
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return -1
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# 尝试解析两行格式
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if len(lines) >= 2:
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count_line = lines[0]
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quant_line = lines[1].upper()
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if count_line.isdigit() and len(quant_line) == int(count_line) and all(c in 'AE' for c in quant_line):
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return (int(count_line), quant_line)
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# 尝试单行格式，例如 "1 EA"
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if len(lines) == 1:
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parts = lines[0].split()
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if len(parts) == 2 and parts[0].isdigit() and len(parts[1]) == int(parts[0]):
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quant = parts[1].upper()
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if all(c in 'AE' for c in quant):
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return (int(parts[0]), quant)
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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# 获取参考解
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ref_sol = cls.reference_solution(identity)
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# 处理无解情况
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if isinstance(ref_sol, int):
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return solution == -1
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# 处理有解情况
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return solution == ref_sol
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@classmethod
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def reference_solution(cls, identity):
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# 完全复制参考代码逻辑
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def toposort(graph):
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n = len(graph)
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res = []
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found = [0]*n
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for i in range(n):
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if found[i]:
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continue
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stack = [i]
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while stack:
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node = stack.pop()
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if node < 0:
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res.append(~node)
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elif not found[node]:
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found[node] = 1
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stack.append(~node)
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for nei in graph[node]:
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if not found[nei]:
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stack.append(nei)
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# Check cycle
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found = [0]*n
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for node in res:
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if found[node]:
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return None
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stack = [node]
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found[node] = 1
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while stack:
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current = stack.pop()
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for nei in graph[current]:
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if found[nei]:
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return None
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if not found[nei]:
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found[nei] = 1
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stack.append(nei)
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return res[::-1]
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n = identity['n']
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edges = identity['edges']
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coupl1 = [[] for _ in range(n)]
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coupl2 = [[] for _ in range(n)]
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for j, k in edges:
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u = j - 1
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v = k - 1
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coupl1[u].append(v)
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coupl2[v].append(u)
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order = toposort(coupl1)
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if order is None:
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return -1
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seen1 = list(range(n))
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seen2 = list(range(n))
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for node in order:
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for nei in coupl1[node]:
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if seen1[nei] > seen1[node]:
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seen1[nei] = seen1[node]
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for node in reversed(order):
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for nei in coupl2[node]:
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if seen2[nei] > seen2[node]:
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seen2[nei] = seen2[node]
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seen = [(seen1[i] == i and seen2[i] == i) for i in range(n)]
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count = sum(seen)
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if count == 0:
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return -1
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quant = ''.join('A' if c else 'E' for c in seen)
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return (count, quant)
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