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* feat(run_eval): add checkpoint resume functionality and update example documentation; - update new bootcamp benchmark dataset * refactor(data_pipeline): optimize data generation pipeline; add multiple preset configurations for data generation * docs: update bootcamp list and add new scripts - Update Fulllist_InternBootcamp.md with new bootcamps and categories - Add new scripts to .gitignore: - examples/pipelines/filter_autogen_configs.py - examples/pipelines/quickgen_data_configs_from_eval_meta.py - Update dependencies in setup.py: - Add scipy and scikit-learn * refactor(internbootcamp): update bootcamp modules and improve error handling - Update import statements in __init__.py files - Add timestamp to target directory name in verl_data_preprocess.py - Improve error handling and scoring logic in bootcamp_judger.py - Remove unnecessary comments and update puzzle descriptions in multiple files
233 lines
6.3 KiB
Python
Executable file
233 lines
6.3 KiB
Python
Executable file
"""### 谜题描述
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Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
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Two matrices A and B are given, each of them has size n × m. Nastya can perform the following operation to matrix A unlimited number of times:
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* take any square square submatrix of A and transpose it (i.e. the element of the submatrix which was in the i-th row and j-th column of the submatrix will be in the j-th row and i-th column after transposing, and the transposed submatrix itself will keep its place in the matrix A).
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Nastya's task is to check whether it is possible to transform the matrix A to the matrix B.
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<image> Example of the operation
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As it may require a lot of operations, you are asked to answer this question for Nastya.
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A square submatrix of matrix M is a matrix which consist of all elements which comes from one of the rows with indeces x, x+1, ..., x+k-1 of matrix M and comes from one of the columns with indeces y, y+1, ..., y+k-1 of matrix M. k is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
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Input
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The first line contains two integers n and m separated by space (1 ≤ n, m ≤ 500) — the numbers of rows and columns in A and B respectively.
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Each of the next n lines contains m integers, the j-th number in the i-th of these lines denotes the j-th element of the i-th row of the matrix A (1 ≤ A_{ij} ≤ 10^{9}).
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Each of the next n lines contains m integers, the j-th number in the i-th of these lines denotes the j-th element of the i-th row of the matrix B (1 ≤ B_{ij} ≤ 10^{9}).
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Output
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Print \"YES\" (without quotes) if it is possible to transform A to B and \"NO\" (without quotes) otherwise.
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You can print each letter in any case (upper or lower).
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Examples
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Input
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2 2
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1 1
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6 1
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1 6
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1 1
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Output
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YES
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Input
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2 2
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4 4
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4 5
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5 4
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4 4
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Output
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NO
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Input
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3 3
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1 2 3
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4 5 6
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7 8 9
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1 4 7
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2 5 6
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3 8 9
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Output
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YES
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Note
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Consider the third example. The matrix A initially looks as follows.
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$$$ \begin{bmatrix} 1 & 2 & 3\\\ 4 & 5 & 6\\\ 7 & 8 & 9 \end{bmatrix} $$$
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Then we choose the whole matrix as transposed submatrix and it becomes
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$$$ \begin{bmatrix} 1 & 4 & 7\\\ 2 & 5 & 8\\\ 3 & 6 & 9 \end{bmatrix} $$$
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Then we transpose the submatrix with corners in cells (2, 2) and (3, 3).
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$$$ \begin{bmatrix} 1 & 4 & 7\\\ 2 & 5 & 8\\\ 3 & 6 & 9 \end{bmatrix} $$$
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So matrix becomes
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$$$ \begin{bmatrix} 1 & 4 & 7\\\ 2 & 5 & 6\\\ 3 & 8 & 9 \end{bmatrix} $$$
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and it is B.
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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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n, m = map(int, raw_input().split())
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a = []
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for i in range(n):
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a.append(map(int, raw_input().split()))
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b = []
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for i in range(n):
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b.append(map(int, raw_input().split()))
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z = 'YES'
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for t in range(n + m - 1):
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if t < n:
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i = t
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j = 0
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else:
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i = n - 1
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j = t - n + 1
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pa = []
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pb = []
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while i >= 0 and j < m:
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pa.append(a[i][j])
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pb.append(b[i][j])
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i -= 1
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j += 1
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pa.sort()
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pb.sort()
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for i in range(len(pa)):
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if pa[i] != pb[i]: z = 'NO'
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if z == 'NO': break
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print z
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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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import re
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from bootcamp import Basebootcamp
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class Cnastyaistransposingmatricesbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.n = params.get('n', 3)
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self.m = params.get('m', 3)
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self.ensure_solvable = params.get('ensure_solvable', True) # 参数命名更明确
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def case_generator(self):
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n, m = self.n, self.m
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B = [[random.randint(1, 100) for _ in range(m)] for _ in range(n)]
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A = [[0]*m for _ in range(n)]
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# 按斜对角线处理元素
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for t in range(n + m -1):
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coords = []
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if t < n:
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i, j = t, 0
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else:
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i, j = n-1, t - n +1
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while i >= 0 and j < m:
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coords.append((i, j))
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i -= 1
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j += 1
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b_elements = [B[x][y] for x, y in coords]
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a_elements = random.sample(b_elements, len(b_elements)) # 确保元素完全重排
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for idx, (x, y) in enumerate(coords):
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A[x][y] = a_elements[idx]
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# 生成错误案例的增强逻辑
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if not self.ensure_solvable:
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diagonal_id = random.randint(0, n + m -2)
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coords = self._get_diagonal_coords(diagonal_id, n, m)
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if coords:
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x, y = random.choice(coords)
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while True:
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new_val = A[x][y] + random.choice([-1, 1])
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if new_val != A[x][y] and new_val > 0:
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A[x][y] = new_val
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break
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return {'n':n, 'm':m, 'A':A, 'B':B}
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@staticmethod
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def _get_diagonal_coords(t, n, m):
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coords = []
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if t < n:
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i, j = t, 0
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else:
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i, j = n-1, t - n +1
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while i >=0 and j < m:
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coords.append((i, j))
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i -= 1
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j += 1
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return coords
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@staticmethod
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def prompt_func(question_case) -> str:
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matrix_text = lambda m: '\n'.join([f"Row {i+1}: {' '.join(map(str, row))}" for i, row in enumerate(m)])
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return f"""Determine if matrix A can be transformed into matrix B using square submatrix transposes.
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Matrix A:
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{matrix_text(question_case['A'])}
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Matrix B:
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{matrix_text(question_case['B'])}
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Answer must be uppercase 'YES' or 'NO' within [answer] tags."""
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@staticmethod
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def extract_output(output):
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answers = re.findall(r'\[answer\](YES|NO)\[/answer\]', output, re.IGNORECASE)
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return answers[-1].upper() if answers else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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for t in range(identity['n'] + identity['m'] -1):
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coords = cls._get_diagonal_coords(t, identity['n'], identity['m'])
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a_values = [identity['A'][x][y] for x, y in coords]
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b_values = [identity['B'][x][y] for x, y in coords]
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if sorted(a_values) != sorted(b_values):
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return solution == 'NO'
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return solution == 'YES'
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