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internbootcamp/bootcamp/equantifierquestion/equantifierquestion.py

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+"""# 
+
+### 谜题描述
+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.
+
+The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:
+
+  * ∀ x,x<100 is read as: for all real numbers x, x is less than 100. This statement is false. 
+  * ∀ x,x>x-1 is read as: for all real numbers x, x is greater than x-1. This statement is true. 
+
+
+
+The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:
+
+  * ∃ x,x<100 is read as: there exists a real number x such that x is less than 100. This statement is true. 
+  * ∃ 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. 
+
+
+
+Moreover, these quantifiers can be nested. For example:
+
+  * ∀ 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. 
+  * ∃ 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. 
+
+
+
+Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.
+
+There are n variables x_1,x_2,…,x_n, and you are given some formula of the form $$$ f(x_1,...,x_n):=(x_{j_1}<x_{k_1})∧ (x_{j_2}<x_{k_2})∧ ⋅⋅⋅∧ (x_{j_m}<x_{k_m}), $$$
+
+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.
+
+Your task is to assign quantifiers Q_1,…,Q_n to either universal (∀) or existential (∃) so that the statement $$$ Q_1 x_1, Q_2 x_2, …, Q_n x_n, f(x_1,…, x_n) $$$
+
+is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.
+
+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.
+
+Input
+
+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.
+
+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).
+
+Output
+
+If there is no assignment of quantifiers for which the statement is true, output a single integer -1.
+
+Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.
+
+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.
+
+Examples
+
+Input
+
+
+2 1
+1 2
+
+
+Output
+
+
+1
+AE
+
+
+Input
+
+
+4 3
+1 2
+2 3
+3 1
+
+
+Output
+
+
+-1
+
+
+Input
+
+
+3 2
+1 3
+2 3
+
+
+Output
+
+
+2
+AAE
+
+Note
+
+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.
+
+For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.
+
+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.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 2e5 + 10;
+const int INF = 0x3f3f3f3f;
+const long long inf = 0x3f3f3f3f3f3f3f3f;
+int n, m;
+vector<int> G[N], GG[N];
+int vis[N], mn[N], mn2[N];
+bool dfs(int u) {
+  vis[u] = 1;
+  mn[u] = u;
+  for (int v : G[u]) {
+    if (vis[v] == 1) return false;
+    if (!vis[v]) {
+      if (!dfs(v)) return false;
+    }
+    mn[u] = min(mn[u], mn[v]);
+  }
+  vis[u] = 2;
+  return true;
+}
+bool dfs2(int u) {
+  vis[u] = 1;
+  mn2[u] = u;
+  for (int v : GG[u]) {
+    if (vis[v] == 1) return false;
+    if (!vis[v]) {
+      if (!dfs2(v)) return false;
+    }
+    mn2[u] = min(mn2[u], mn2[v]);
+  }
+  vis[u] = 2;
+  return true;
+}
+int main() {
+  scanf(\"%d%d\", &n, &m);
+  for (int i = 1; i <= m; i++) {
+    int u, v;
+    scanf(\"%d%d\", &u, &v);
+    G[v].push_back(u);
+    GG[u].push_back(v);
+  }
+  for (int i = 1; i <= n; i++)
+    if (!vis[i]) {
+      if (!dfs(i)) {
+        puts(\"-1\");
+        return 0;
+      }
+    }
+  memset(vis, 0, sizeof(vis));
+  for (int i = 1; i <= n; i++)
+    if (!vis[i]) {
+      if (!dfs2(i)) {
+        puts(\"-1\");
+        return 0;
+      }
+    }
+  int ans = 0;
+  for (int i = 1; i <= n; i++)
+    if (mn[i] >= i && mn2[i] >= i) ans++;
+  printf(\"%d\n\", ans);
+  for (int i = 1; i <= n; i++)
+    printf(\"%c\", mn[i] >= i && mn2[i] >= i ? 'A' : 'E');
+  puts(\"\");
+  return 0;
+}
+```
+
+
+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。
+"""
+
+from bootcamp import Basebootcamp
+import random
+from bootcamp import Basebootcamp
+
+class Equantifierquestionbootcamp(Basebootcamp):
+    def __init__(self, n_min=2, n_max=10, m_min=1, m_max=20):
+        self.n_min = n_min
+        self.n_max = n_max
+        self.m_min = m_min
+        self.m_max = m_max
+    
+    def case_generator(self):
+        n = random.randint(self.n_min, self.n_max)
+        m = random.randint(self.m_min, min(self.m_max, n * (n - 1)))
+        
+        inequalities = []
+        for _ in range(m):
+            j = random.randint(1, n)
+            k = random.randint(1, n)
+            while j == k:
+                k = random.randint(1, n)
+            inequalities.append((j, k))
+        
+        G = [[] for _ in range(n + 1)]
+        GG = [[] for _ in range(n + 1)]
+        for j, k in inequalities:
+            G[k].append(j)
+            GG[j].append(k)
+        
+        vis = [0] * (n + 1)
+        mn = [0] * (n + 1)
+        has_cycle = False
+        
+        def dfs(u):
+            nonlocal has_cycle
+            if vis[u] == 1:
+                has_cycle = True
+                return
+            if vis[u] == 2:
+                return
+            vis[u] = 1
+            mn_u = u
+            for v in G[u]:
+                dfs(v)
+                if has_cycle:
+                    return
+                mn_u = min(mn_u, mn[v])
+            mn[u] = mn_u
+            vis[u] = 2
+        
+        for i in range(1, n + 1):
+            if vis[i] == 0:
+                dfs(i)
+                if has_cycle:
+                    break
+        
+        if has_cycle:
+            return {'n': n, 'm': m, 'inequalities': inequalities, 'answer': -1}
+        
+        vis2 = [0] * (n + 1)
+        mn2 = [0] * (n + 1)
+        has_cycle2 = False
+        
+        def dfs2(u):
+            nonlocal has_cycle2
+            if vis2[u] == 1:
+                has_cycle2 = True
+                return
+            if vis2[u] == 2:
+                return
+            vis2[u] = 1
+            mn2_u = u
+            for v in GG[u]:
+                dfs2(v)
+                if has_cycle2:
+                    return
+                mn2_u = min(mn2_u, mn2[v])
+            mn2[u] = mn2_u
+            vis2[u] = 2
+        
+        for i in range(1, n + 1):
+            if vis2[i] == 0:
+                dfs2(i)
+                if has_cycle2:
+                    break
+        
+        if has_cycle2:
+            return {'n': n, 'm': m, 'inequalities': inequalities, 'answer': -1}
+        
+        ans = []
+        max_universal = 0
+        for i in range(1, n + 1):
+            if mn[i] >= i and mn2[i] >= i:
+                ans.append('A')
+                max_universal += 1
+            else:
+                ans.append('E')
+        ans_str = ''.join(ans)
+        
+        return {
+            'n': n,
+            'm': m,
+            'inequalities': inequalities,
+            'answer': ans_str,
+            'max_universal': max_universal
+        }
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        m = question_case['m']
+        inequalities = question_case['inequalities']
+        
+        ineq_text = []
+        for j, k in inequalities:
+            ineq_text.append(f"x_{j} < x_{k}")
+        ineq_text_str = " ∧ ".join(ineq_text)
+        
+        prompt = f"""
+        你是一位逻辑学专家，现在需要解决一个关于量词分配的问题。给定{n}个变量和{m}个不等式，每个不等式是x_j < x_k的形式。你的任务是为每个变量分配量词∀或∃，使得整个命题为真，并且尽可能多地使用∀。如果无解，输出-1。
+
+        具体的不等式如下：
+        f(x_1, ..., x_{n}) = {ineq_text_str}
+
+        请输出你的解决方案，将所有变量的量词分配写在一个字符串中，例如'AAE'，并放在[answer][/answer]标签中。
+        """
+        return prompt.strip()
+    
+    @staticmethod
+    def extract_output(output):
+        start = output.rfind('[answer]')
+        if start == -1:
+            return None
+        end = output.find('[/answer]', start)
+        if end == -1:
+            return None
+        answer = output[start + len('[answer]'):end].strip().upper()
+        return answer
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        if solution is None:
+            return False
+        if len(solution) != identity['n']:
+            return False
+        if not all(c in {'A', 'E'} for c in solution):
+            return False
+        return solution == identity['answer']