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internbootcamp/bootcamp/cbarcode/cbarcode.py

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+"""# 
+
+### 谜题描述
+You've got an n × m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
+
+A picture is a barcode if the following conditions are fulfilled: 
+
+  * All pixels in each column are of the same color. 
+  * The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y. 
+
+Input
+
+The first line contains four space-separated integers n, m, x and y (1 ≤ n, m, x, y ≤ 1000; x ≤ y).
+
+Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character \".\" represents a white pixel and \"#\" represents a black pixel. The picture description doesn't have any other characters besides \".\" and \"#\".
+
+Output
+
+In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists. 
+
+Examples
+
+Input
+
+6 5 1 2
+##.#.
+.###.
+###..
+#...#
+.##.#
+###..
+
+
+Output
+
+11
+
+
+Input
+
+2 5 1 1
+#####
+.....
+
+
+Output
+
+5
+
+Note
+
+In the first test sample the picture after changing some colors can looks as follows: 
+    
+    
+      
+    .##..  
+    .##..  
+    .##..  
+    .##..  
+    .##..  
+    .##..  
+    
+
+In the second test sample the picture after changing some colors can looks as follows: 
+    
+    
+      
+    .#.#.  
+    .#.#.  
+    
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+n, m, x, y = map(int, raw_input().split())
+u = [i.count('.') for i in list(zip(*[raw_input() for i in range(n)]))]
+v = [n - i for i in u]
+a, b, s = [u[0]], [v[0]], x - 1
+for i in range(1, x):
+    a, b = [1000001] + [j + u[i] for j in a], [1000001] + [j + v[i] for j in b]
+for i in range(x, min(y, m)):
+    a, b = [min(b[s: ]) + u[i]] + [j + u[i] for j in a], [min(a[s: ]) + v[i]] + [j + v[i] for j in b]
+for i in range(min(y, m), m):
+    a, b = [min(b[s: ]) + u[i]] + [j + u[i] for j in a[: -1]], [min(a[s: ]) + v[i]] + [j + v[i] for j in b[: -1]]
+print(min(min(a[s: ]), min(b[s: ])))
+```
+
+
+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
+
+def calculate_min_pixels(n, m, x, y, rows):
+    # 转置行以获取各列
+    cols = list(zip(*rows))
+    u = [col.count('.') for col in cols]  # 每列改为白色所需的修改次数（即列中黑色像素数）
+    v = [n - count for count in u]        # 每列改为黑色所需的修改次数（即列中白色像素数）
+    
+    a = [u[0]]  # 以白色结尾的段的最小修改次数
+    b = [v[0]]  # 以黑色结尾的段的最小修改次数
+    s = x - 1   # 段长度至少为x，因此需要保留前s个状态
+    
+    # 处理前x-1列
+    for i in range(1, x):
+        # 由于段长度必须>=x，此时只能继续延长当前颜色段
+        a = [float('inf')] + [prev + u[i] for prev in a]
+        b = [float('inf')] + [prev + v[i] for prev in b]
+    
+    # 处理x到min(y, m)-1列
+    for i in range(x, min(y, m)):
+        # 可以开始新的颜色段，此时需要取另一种颜色的最小值
+        min_b = min(b[s:]) if b[s:] else float('inf')
+        new_a = [min_b + u[i]] + [prev + u[i] for prev in a]
+        min_a = min(a[s:]) if a[s:] else float('inf')
+        new_b = [min_a + v[i]] + [prev + v[i] for prev in b]
+        a, b = new_a, new_b
+    
+    # 处理剩下的列（当m > y时）
+    for i in range(min(y, m), m):
+        # 需要确保段长度不超过y，因此保留前y个状态
+        min_b = min(b[s:]) if b[s:] else float('inf')
+        new_a = [min_b + u[i]] + [prev + u[i] for prev in a[:-1]]  # 保留前y-1个状态
+        min_a = min(a[s:]) if a[s:] else float('inf')
+        new_b = [min_a + v[i]] + [prev + v[i] for prev in b[:-1]]
+        a, b = new_a, new_b
+    
+    # 最后，取所有可能状态中的最小值
+    valid_a = a[s:] if a[s:] else [float('inf')]
+    valid_b = b[s:] if b[s:] else [float('inf')]
+    return min(min(valid_a), min(valid_b))
+
+class Cbarcodebootcamp(Basebootcamp):
+    def __init__(self, n=6, m=5, x=1, y=2):
+        self.n = n
+        self.m = m
+        self.x = min(x, y)
+        self.y = max(x, y)
+        
+    def case_generator(self):
+        # 生成随机图像
+        rows = []
+        for _ in range(self.n):
+            row = ''.join(random.choice(['#', '.']) for _ in range(self.m))
+            rows.append(row)
+        
+        # 计算正确答案
+        correct_answer = calculate_min_pixels(self.n, self.m, self.x, self.y, rows)
+        
+        return {
+            'n': self.n,
+            'm': self.m,
+            'x': self.x,
+            'y': self.y,
+            'rows': rows,
+            'correct_answer': correct_answer
+        }
+    
+    @staticmethod
+    def prompt_func(question_case):
+        prompt = [
+            "You are a barcode conversion expert. Your task is to transform the given image into a valid barcode by repainting as few pixels as possible.",
+            "",
+            "A valid barcode must satisfy:",
+            "1. Each column consists of entirely black (#) or white (.) pixels.",
+            "2. The width of every consecutive vertical line (group of same-color columns) must be between x and y, inclusive.",
+            "",
+            f"Problem parameters:",
+            f"- Rows (n): {question_case['n']}",
+            f"- Columns (m): {question_case['m']}",
+            f"- Minimum width (x): {question_case['x']}",
+            f"- Maximum width (y): {question_case['y']}",
+            "",
+            "Original image (each line represents a row):"
+        ]
+        prompt.extend(question_case['rows'])
+        prompt.append(
+            "\nCalculate the minimal number of pixels to repaint. Put your final answer within [answer] and [/answer]."
+        )
+        return '\n'.join(prompt)
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
+        if not matches:
+            return None
+        try:
+            return int(matches[-1].strip())
+        except:
+            return None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        return solution == identity.get('correct_answer', None)