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259 lines
8.1 KiB
Python
Executable file
259 lines
8.1 KiB
Python
Executable file
"""#
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### 谜题描述
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This is an interactive problem
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You are given a grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
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Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
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We want to know numbers in all cells of the grid. To do so we can ask the following questions:
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\"? x_1 y_1 x_2 y_2\", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
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As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
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For example, paths, shown in green, are palindromic, so answer for \"? 1 1 2 3\" and \"? 1 2 3 3\" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
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<image>
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Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
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Input
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The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
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Interaction
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You begin the interaction by reading n.
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To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output \"? x_1 y_1 x_2 y_2\".
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Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
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In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
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In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
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When you determine numbers in all cells, output \"!\".
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Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
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After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
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* fflush(stdout) or cout.flush() in C++;
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* System.out.flush() in Java;
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* flush(output) in Pascal;
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* stdout.flush() in Python;
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* see documentation for other languages.
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Hack Format
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To hack, use the following format.
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The first line should contain a single odd integer n (side of your grid).
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The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
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Example
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Input
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3
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0
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1
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0
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1
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1
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1
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1
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Output
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? 1 1 1 3
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? 1 1 2 3
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? 2 1 2 3
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? 3 1 3 3
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? 2 2 3 3
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? 1 2 3 2
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? 1 2 3 3
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!
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100
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001
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000
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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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from sys import stdout
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def Q(x1, y1, x2, y2):
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print '?', x1 + 1, y1 + 1, x2 + 1, y2 + 1
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stdout.flush()
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return raw_input()[0] == '1'
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def A(a):
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n = len(a)
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print \"!\"
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for i in xrange(n):
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print ''.join(map(str, a[i]))
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stdout.flush()
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quit()
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def q(x1, y1, x2, y2, a, b, rev=0):
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x = Q(x1, y1, x2, y2) ^ 1
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if rev:
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x1, y1, x2, y2 = x2, y2, x1, y1
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a[x2][y2] = a[x1][y1] ^ x
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b[x2][y2] = b[x1][y1] ^ x
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def ok(a, x1, y1, x2, y2):
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x = Q(x1, y1, x2, y2)
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s = {}
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f = 0
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if x2 - x1 == 1:
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s[y1] = str(a[x1][y1])
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for i in xrange(y1 + 1, y2 + 1):
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s[i] = s[i-1] + str(a[x1][i])
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for i in xrange(y1, y2 + 1):
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t = s[i]
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for j in xrange(i, y2 + 1):
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t += str(a[x1+1][j])
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if t == t[::-1]:
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f = 1
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else:
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s[x1] = str(a[x1][y1])
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for i in xrange(x1 + 1, x2 + 1):
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s[i] = s[i-1] + str(a[i][y1])
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for i in xrange(x1, x2 + 1):
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t = s[i]
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for j in xrange(i, x2 + 1):
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t += str(a[j][y1+1])
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if t == t[::-1]:
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f = 1
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return x == f
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def main():
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n = int(raw_input())
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a = [[-1] * n for _ in xrange(n)]
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a[0][0] = 1
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a[-1][-1] = 0
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b = [l[:] for l in a]
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a[0][1] = 0
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b[0][1] = 1
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q(0, 0, 1, 1, a, b)
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for i in xrange(n):
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j = i % 2
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if i >= 2:
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q(i - 2, j, i, j, a, b)
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nd = n
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if i == n - 1:
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nd -= 2
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while j + 2 < nd:
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q(i, j, i, j + 2, a, b)
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j += 2
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for i in xrange(1, n - 3, 2):
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q(0, i, 0, i + 2, a, b)
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q(0, n - 2, 1, n - 1, a, b)
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for i in xrange(n - 1, 1, -2):
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q(1, i - 2, 1, i, a, b, 1)
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for i in xrange(2, n):
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j = (i % 2) ^ 1
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q(i - 2, j, i, j, a, b)
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while j + 2 < n:
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q(i, j, i, j + 2, a, b)
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j += 2
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for i in xrange(0, n - 1, 2):
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for j in xrange(0, n - 1, 2):
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if a[i][j] != a[i+2][j+2]:
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if ok(a, i, j, i + 1, j + 2) and ok(a, i, j, i + 2, j + 1) and ok(a, i + 1, j, i + 2, j + 2):# and ok(a, i, j + 1, i + 2, j + 2):
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A(a)
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else:
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A(b)
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main()
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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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class Cpalindromicpathsbootcamp(Basebootcamp):
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def __init__(self, n=3):
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if (n % 2) == 0:
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n += 1 # 确保n为奇数
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self.n = n
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def case_generator(self):
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n = self.n
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grid = [[0] * n for _ in range(n)]
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grid[0][0] = 1 # 左上角为1
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grid[-1][-1] = 0 # 右下角为0
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# 随机生成其他单元格的值
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for i in range(n):
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for j in range(n):
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if i == 0 and j == 0:
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continue
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if i == n-1 and j == n-1:
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continue
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grid[i][j] = random.choice([0, 1])
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return {
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'n': n,
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'grid': grid
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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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grid = question_case['grid']
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example = "\n".join(["".join(map(str, row)) for row in grid])
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prompt = f"你有一个{n}×{n}的网格。已知左上角（1,1）的值是1，右下角（{n},{n}）的值是0。其他单元格的值是0或1。你需要通过询问问题来确定整个网格的值。每次询问的形式是“? x1 y1 x2 y2”，并得到0或1的回答。你的任务是输出整个网格的值，每行是一个由0和1组成的字符串，例如：\n\n"
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prompt += f"示例网格：\n{example}\n\n"
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prompt += "请将答案放在[answer]标签中，格式如下：\n"
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prompt += "[answer]\n"
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prompt += "100\n"
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prompt += "001\n"
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prompt += "000\n"
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prompt += "[/answer]\n"
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return prompt
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@staticmethod
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def extract_output(output):
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import re
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pattern = r'\[answer\](.*?)\[\/answer\]'
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matches = re.findall(pattern, output, re.DOTALL)
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if not matches:
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return None
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answer = matches[-1].strip()
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rows = answer.split('\n')
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rows = [row.strip() for row in rows if row.strip() != '']
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for row in rows:
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if not all(c in '01' for c in row):
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return None
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return rows
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@classmethod
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def _verify_correction(cls, solution, identity):
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n = identity['n']
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expected_grid = identity['grid']
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if len(solution) != n:
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return False
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for row in solution:
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if len(row) != n:
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return False
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actual_grid = []
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for row in solution:
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actual_grid.append([int(c) for c in row])
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for i in range(n):
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for j in range(n):
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if actual_grid[i][j] != expected_grid[i][j]:
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return False
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return True
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