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327 lines
9.9 KiB
Python
Executable file
327 lines
9.9 KiB
Python
Executable file
"""#
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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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#include <bits/stdc++.h>
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using namespace std;
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const int N = 2e5 + 10;
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const int INF = 0x3f3f3f3f;
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const long long inf = 0x3f3f3f3f3f3f3f3f;
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int n, m;
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vector<int> G[N], GG[N];
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int vis[N], mn[N], mn2[N];
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bool dfs(int u) {
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vis[u] = 1;
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mn[u] = u;
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for (int v : G[u]) {
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if (vis[v] == 1) return false;
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if (!vis[v]) {
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if (!dfs(v)) return false;
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}
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mn[u] = min(mn[u], mn[v]);
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}
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vis[u] = 2;
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return true;
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}
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bool dfs2(int u) {
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vis[u] = 1;
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mn2[u] = u;
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for (int v : GG[u]) {
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if (vis[v] == 1) return false;
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if (!vis[v]) {
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if (!dfs2(v)) return false;
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}
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mn2[u] = min(mn2[u], mn2[v]);
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}
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vis[u] = 2;
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return true;
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}
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int main() {
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scanf(\"%d%d\", &n, &m);
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for (int i = 1; i <= m; i++) {
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int u, v;
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scanf(\"%d%d\", &u, &v);
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G[v].push_back(u);
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GG[u].push_back(v);
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}
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for (int i = 1; i <= n; i++)
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if (!vis[i]) {
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if (!dfs(i)) {
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puts(\"-1\");
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return 0;
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}
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}
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memset(vis, 0, sizeof(vis));
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for (int i = 1; i <= n; i++)
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if (!vis[i]) {
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if (!dfs2(i)) {
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puts(\"-1\");
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return 0;
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}
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}
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int ans = 0;
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for (int i = 1; i <= n; i++)
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if (mn[i] >= i && mn2[i] >= i) ans++;
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printf(\"%d\n\", ans);
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for (int i = 1; i <= n; i++)
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printf(\"%c\", mn[i] >= i && mn2[i] >= i ? 'A' : 'E');
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puts(\"\");
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return 0;
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}
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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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import random
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from bootcamp import Basebootcamp
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class Equantifierquestionbootcamp(Basebootcamp):
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def __init__(self, n_min=2, n_max=10, m_min=1, m_max=20):
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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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n = random.randint(self.n_min, self.n_max)
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m = random.randint(self.m_min, min(self.m_max, n * (n - 1)))
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inequalities = []
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for _ in range(m):
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j = random.randint(1, n)
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k = random.randint(1, n)
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while j == k:
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k = random.randint(1, n)
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inequalities.append((j, k))
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G = [[] for _ in range(n + 1)]
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GG = [[] for _ in range(n + 1)]
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for j, k in inequalities:
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G[k].append(j)
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GG[j].append(k)
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vis = [0] * (n + 1)
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mn = [0] * (n + 1)
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has_cycle = False
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def dfs(u):
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nonlocal has_cycle
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if vis[u] == 1:
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has_cycle = True
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return
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if vis[u] == 2:
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return
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vis[u] = 1
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mn_u = u
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for v in G[u]:
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dfs(v)
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if has_cycle:
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return
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mn_u = min(mn_u, mn[v])
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mn[u] = mn_u
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vis[u] = 2
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for i in range(1, n + 1):
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if vis[i] == 0:
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dfs(i)
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if has_cycle:
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break
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if has_cycle:
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return {'n': n, 'm': m, 'inequalities': inequalities, 'answer': -1}
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vis2 = [0] * (n + 1)
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mn2 = [0] * (n + 1)
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has_cycle2 = False
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def dfs2(u):
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nonlocal has_cycle2
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if vis2[u] == 1:
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has_cycle2 = True
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return
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if vis2[u] == 2:
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return
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vis2[u] = 1
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mn2_u = u
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for v in GG[u]:
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dfs2(v)
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if has_cycle2:
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return
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mn2_u = min(mn2_u, mn2[v])
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mn2[u] = mn2_u
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vis2[u] = 2
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for i in range(1, n + 1):
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if vis2[i] == 0:
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dfs2(i)
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if has_cycle2:
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break
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if has_cycle2:
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return {'n': n, 'm': m, 'inequalities': inequalities, 'answer': -1}
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ans = []
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max_universal = 0
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for i in range(1, n + 1):
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if mn[i] >= i and mn2[i] >= i:
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ans.append('A')
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max_universal += 1
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else:
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ans.append('E')
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ans_str = ''.join(ans)
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return {
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'n': n,
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'm': m,
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'inequalities': inequalities,
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'answer': ans_str,
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'max_universal': max_universal
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}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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m = question_case['m']
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inequalities = question_case['inequalities']
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ineq_text = []
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for j, k in inequalities:
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ineq_text.append(f"x_{j} < x_{k}")
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ineq_text_str = " ∧ ".join(ineq_text)
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prompt = f"""
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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。
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具体的不等式如下：
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f(x_1, ..., x_{n}) = {ineq_text_str}
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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。
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"""
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return prompt.strip()
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@staticmethod
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def extract_output(output):
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start = output.rfind('[answer]')
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if start == -1:
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return None
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end = output.find('[/answer]', start)
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if end == -1:
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return None
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answer = output[start + len('[answer]'):end].strip().upper()
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return answer
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@classmethod
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def _verify_correction(cls, solution, identity):
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if solution is None:
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return False
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if len(solution) != identity['n']:
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return False
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if not all(c in {'A', 'E'} for c in solution):
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return False
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return solution == identity['answer']
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