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internbootcamp/bootcamp/dunmerge/dunmerge.py

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+"""# 
+
+### 谜题描述
+Let a and b be two arrays of lengths n and m, respectively, with no elements in common. We can define a new array merge(a,b) of length n+m recursively as follows:
+
+  * If one of the arrays is empty, the result is the other array. That is, merge(∅,b)=b and merge(a,∅)=a. In particular, merge(∅,∅)=∅. 
+  * If both arrays are non-empty, and a_1<b_1, then merge(a,b)=[a_1]+merge([a_2,…,a_n],b). That is, we delete the first element a_1 of a, merge the remaining arrays, then add a_1 to the beginning of the result. 
+  * If both arrays are non-empty, and a_1>b_1, then merge(a,b)=[b_1]+merge(a,[b_2,…,b_m]). That is, we delete the first element b_1 of b, merge the remaining arrays, then add b_1 to the beginning of the result. 
+
+
+
+This algorithm has the nice property that if a and b are sorted, then merge(a,b) will also be sorted. For example, it is used as a subroutine in merge-sort. For this problem, however, we will consider the same procedure acting on non-sorted arrays as well. For example, if a=[3,1] and b=[2,4], then merge(a,b)=[2,3,1,4].
+
+A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
+
+There is a permutation p of length 2n. Determine if there exist two arrays a and b, each of length n and with no elements in common, so that p=merge(a,b).
+
+Input
+
+The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases. 
+
+The first line of each test case contains a single integer n (1≤ n≤ 2000).
+
+The second line of each test case contains 2n integers p_1,…,p_{2n} (1≤ p_i≤ 2n). It is guaranteed that p is a permutation.
+
+It is guaranteed that the sum of n across all test cases does not exceed 2000.
+
+Output
+
+For each test case, output \"YES\" if there exist arrays a, b, each of length n and with no common elements, so that p=merge(a,b). Otherwise, output \"NO\".
+
+Example
+
+Input
+
+
+6
+2
+2 3 1 4
+2
+3 1 2 4
+4
+3 2 6 1 5 7 8 4
+3
+1 2 3 4 5 6
+4
+6 1 3 7 4 5 8 2
+6
+4 3 2 5 1 11 9 12 8 6 10 7
+
+
+Output
+
+
+YES
+NO
+YES
+YES
+NO
+NO
+
+Note
+
+In the first test case, [2,3,1,4]=merge([3,1],[2,4]).
+
+In the second test case, we can show that [3,1,2,4] is not the merge of two arrays of length 2.
+
+In the third test case, [3,2,6,1,5,7,8,4]=merge([3,2,8,4],[6,1,5,7]).
+
+In the fourth test case, [1,2,3,4,5,6]=merge([1,3,6],[2,4,5]), for example.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+from __future__ import division, print_function
+_interactive = False
+
+def main():
+    for _ in range(int(input())):
+        n = int(input())
+        ar = input_as_list()
+
+        blocks = []
+        prev = ar[0]
+        sz = 0
+        for x in ar:
+            if x > prev:
+                prev = x
+                blocks += [sz]
+                sz = 0
+            sz += 1
+        blocks += [sz]
+
+        dp = [1] + [0]*n
+        for b in blocks:
+            for i in reversed(range(n+1)):
+                if dp[i] and i+b <= n:
+                    dp[i+b] = 1
+
+        print(\"YES\" if dp[n] else \"NO\")
+
+
+# Constants
+INF = float('inf')
+MOD = 10**9+7
+alphabets = 'abcdefghijklmnopqrstuvwxyz'
+
+# Python3 equivalent names
+import os, sys, itertools
+if sys.version_info[0] < 3:
+    input = raw_input
+    range = xrange
+
+    filter = itertools.ifilter
+    map = itertools.imap
+    zip = itertools.izip
+
+# print-flush in interactive problems
+if _interactive:
+    flush = sys.stdout.flush
+    def printf(*args, **kwargs):
+        print(*args, **kwargs)
+        flush()
+
+# Debug print, only works on local machine
+LOCAL = \"LOCAL_\" in os.environ
+debug_print = (print) if LOCAL else (lambda *x, **y: None)
+
+# Fast IO
+if (not LOCAL) and (not _interactive):
+    from io import BytesIO
+    from atexit import register
+    sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
+    sys.stdout = BytesIO()
+    register(lambda: os.write(1, sys.stdout.getvalue()))
+    input = lambda: sys.stdin.readline().rstrip('\r\n')
+
+# Some utility functions(Input, N-dimensional lists, ...)
+def input_as_list():
+    return [int(x) for x in input().split()]
+
+def input_with_offset(o):
+    return [int(x)+o for x in input().split()]
+
+def input_as_matrix(n, m):
+    return [input_as_list() for _ in range(n)]
+
+def array_of(f, *dim):
+    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
+
+main()
+```
+
+
+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
+import re
+from bootcamp import Basebootcamp
+
+class Dunmergebootcamp(Basebootcamp):
+    def __init__(self, **params):
+        super().__init__(**params)
+        self.params = {
+            'min_n': params.get('min_n', 1),
+            'max_n': params.get('max_n', 2000)
+        }
+    
+    def case_generator(self):
+        n = random.randint(self.params['min_n'], self.params['max_n'])
+        all_elements = list(range(1, 2 * n + 1))
+        random.shuffle(all_elements)
+        a = all_elements[:n]
+        b = all_elements[n:]
+        random.shuffle(a)
+        random.shuffle(b)
+        p = self.merge(a, b)
+        return {'n': n, 'p': p}
+    
+    @staticmethod
+    def merge(a, b):
+        result = []
+        i = j = 0
+        while i < len(a) and j < len(b):
+            if a[i] < b[j]:
+                result.append(a[i])
+                i += 1
+            else:
+                result.append(b[j])
+                j += 1
+        result.extend(a[i:])
+        result.extend(b[j:])
+        return result
+    
+    @staticmethod
+    def prompt_func(question_case):
+        n = question_case['n']
+        p = question_case['p']
+        p_str = ' '.join(map(str, p))
+        prompt = f"""
+        你有一个排列p，长度为2n，其中n={n}。p的具体值为：{p_str}。
+
+        你需要判断是否存在两个长度为n的数组a和b，且它们的元素互不相同，使得p可以通过merge(a,b)过程得到。
+
+        merge(a,b)的定义如下：
+        - 如果其中一个数组为空，则结果是另一个数组。
+        - 如果两个数组都不为空，比较a的第一个元素和b的第一个元素，较小的放在前面，然后递归处理剩下的部分。
+
+        例如，a=[3,1]，b=[2,4]，则merge(a,b)=[2,3,1,4]。
+
+        请判断是否存在这样的a和b，并将你的答案（YES或NO）放在[answer]标签中。
+
+        请将答案以以下格式输出：
+        [answer]
+        YES或NO
+        [/answer]
+        """
+        return prompt
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        solution = matches[-1].strip().upper()
+        if solution not in ('YES', 'NO'):
+            return None
+        return solution
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        n = identity['n']
+        p = identity['p']
+        if not p:
+            return False
+        prev = p[0]
+        sz = 0
+        blocks = []
+        for x in p:
+            if x > prev:
+                blocks.append(sz)
+                sz = 0
+                prev = x
+            sz += 1
+        blocks.append(sz)
+        dp = [False] * (n + 1)
+        dp[0] = True
+        for b in blocks:
+            for i in range(n, -1, -1):
+                if dp[i] and (i + b) <= n:
+                    if not dp[i + b]:
+                        dp[i + b] = True
+        expected_answer = 'YES' if dp[n] else 'NO'
+        return solution.upper() == expected_answer