init-commit

internbootcamp/bootcamp/cdesigntutorialmakeitnondeterministic/cdesigntutorialmakeitnondeterministic.py

@@ -0,0 +1,285 @@
+"""# 
+
+### 谜题描述
+A way to make a new task is to make it nondeterministic or probabilistic. For example, the hard task of Topcoder SRM 595, Constellation, is the probabilistic version of a convex hull.
+
+Let's try to make a new task. Firstly we will use the following task. There are n people, sort them by their name. It is just an ordinary sorting problem, but we can make it more interesting by adding nondeterministic element. There are n people, each person will use either his/her first name or last name as a handle. Can the lexicographical order of the handles be exactly equal to the given permutation p?
+
+More formally, if we denote the handle of the i-th person as hi, then the following condition must hold: <image>.
+
+Input
+
+The first line contains an integer n (1 ≤ n ≤ 105) — the number of people.
+
+The next n lines each contains two strings. The i-th line contains strings fi and si (1 ≤ |fi|, |si| ≤ 50) — the first name and last name of the i-th person. Each string consists only of lowercase English letters. All of the given 2n strings will be distinct.
+
+The next line contains n distinct integers: p1, p2, ..., pn (1 ≤ pi ≤ n).
+
+Output
+
+If it is possible, output \"YES\", otherwise output \"NO\".
+
+Examples
+
+Input
+
+3
+gennady korotkevich
+petr mitrichev
+gaoyuan chen
+1 2 3
+
+
+Output
+
+NO
+
+
+Input
+
+3
+gennady korotkevich
+petr mitrichev
+gaoyuan chen
+3 1 2
+
+
+Output
+
+YES
+
+
+Input
+
+2
+galileo galilei
+nicolaus copernicus
+2 1
+
+
+Output
+
+YES
+
+
+Input
+
+10
+rean schwarzer
+fei claussell
+alisa reinford
+eliot craig
+laura arseid
+jusis albarea
+machias regnitz
+sara valestin
+emma millstein
+gaius worzel
+1 2 3 4 5 6 7 8 9 10
+
+
+Output
+
+NO
+
+
+Input
+
+10
+rean schwarzer
+fei claussell
+alisa reinford
+eliot craig
+laura arseid
+jusis albarea
+machias regnitz
+sara valestin
+emma millstein
+gaius worzel
+2 4 9 6 5 7 1 3 8 10
+
+
+Output
+
+YES
+
+Note
+
+In example 1 and 2, we have 3 people: tourist, Petr and me (cgy4ever). You can see that whatever handle is chosen, I must be the first, then tourist and Petr must be the last.
+
+In example 3, if Copernicus uses \"copernicus\" as his handle, everything will be alright.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+def MIN(t):
+    if t[0] < t[1]:
+        return t[0]
+    else:
+        return t[1]
+def MAX(t):
+    if t[0] > t[1]:
+        return t[0]
+    else:
+        return t[1] 
+n = input()
+
+name = []
+while n:
+    n -= 1
+    a,b = raw_input().split()
+    name.append([a,b])
+p = map(int,raw_input().split())
+
+Flag = True
+Tmp = MIN(name[p[0]-1])
+for i in xrange(1,len(p)):
+    if Tmp >= MIN(name[p[i]-1]):
+        if Tmp >= MAX(name[p[i]-1]):
+            Flag = False
+            break
+        else:
+            Tmp = MAX(name[p[i]-1])
+    else:
+        Tmp = MIN(name[p[i]-1])
+
+if Flag == True:
+    print \"YES\"
+else:
+    print \"NO\"
+```
+
+
+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 solve_handle_order(names, p_list):
+    if not names or len(p_list) != len(names):
+        return "NO"
+    n = len(names)
+    p = p_list
+    Flag = True
+    current_user = names[p[0]-1]
+    a, b = current_user
+    Tmp = a if a < b else b
+    for i in range(1, len(p)):
+        current_user = names[p[i]-1]
+        a, b = current_user
+        current_min = a if a < b else b
+        current_max = b if a < b else a
+        if Tmp >= current_min:
+            if Tmp >= current_max:
+                Flag = False
+                break
+            else:
+                Tmp = current_max
+        else:
+            Tmp = current_min
+    return "YES" if Flag else "NO"
+
+def generate_random_string(min_length=1, max_length=50):
+    length = random.randint(min_length, max_length)
+    return ''.join(random.choice('abcdefghijklmnopqrstuvwxyz') for _ in range(length))
+
+class Cdesigntutorialmakeitnondeterministicbootcamp(Basebootcamp):
+    def __init__(self, min_n=3, max_n=10, solvable=None, max_attempts=1000):
+        super().__init__()  # 显式调用父类初始化
+        self.min_n = min_n
+        self.max_n = max_n
+        self.solvable = solvable
+        self.max_attempts = max_attempts
+    
+    def case_generator(self):
+        for _ in range(self.max_attempts):
+            # 保证n至少为1
+            n = random.randint(max(1, self.min_n), self.max_n)
+            
+            names = []
+            all_names = set()
+            valid = True
+            
+            # 为每个用户生成唯一的名字对
+            for _ in range(n):
+                attempt = 0
+                while True:
+                    first = generate_random_string(1, 50)
+                    last = generate_random_string(1, 50)
+                    if first != last and first not in all_names and last not in all_names:
+                        all_names.update([first, last])
+                        names.append((first, last))
+                        break
+                    attempt += 1
+                    if attempt > 100:
+                        valid = False
+                        break
+                if not valid:
+                    break
+            if not valid:
+                continue
+            
+            # 生成有效排列p
+            p = list(range(1, n+1))
+            random.shuffle(p)
+            answer = solve_handle_order(names, p)
+            
+            # 根据solvable参数筛选案例
+            if self.solvable is None:
+                return {'n': n, 'names': names, 'p': p}
+            elif (self.solvable and answer == "YES") or (not self.solvable and answer == "NO"):
+                return {'n': n, 'names': names, 'p': p}
+        
+        # 备用方案：生成确定性的可解/不可解案例
+        if self.solvable:
+            return {'n': 3, 'names': [('a','z'), ('b','y'), ('c','x')], 'p': [3,2,1]}
+        else:
+            return {'n': 3, 'names': [('z','a'), ('y','b'), ('x','c')], 'p': [1,2,3]}
+    
+    @staticmethod
+    def prompt_func(question_case):
+        input_lines = [str(question_case['n'])]
+        for first, last in question_case['names']:
+            input_lines.append(f"{first} {last}")
+        input_lines.append(' '.join(map(str, question_case['p'])))
+        example_input = '\n'.join(input_lines)
+        
+        prompt = (
+            "给定n个人，每个人可以选择使用名字或姓氏作为handle。\n"
+            "判断是否存在一种选择方式，使得按handle的字典序结果恰好等于给定的排列p。\n\n"
+            "输入格式：\n"
+            f"- 首行：n（人数）\n"
+            f"- 接下来n行：每行两个小写字母组成的字符串\n"
+            f"- 最后一行：排列p（1-based索引）\n\n"
+            "示例：\n"
+            "输入：\n3\na b\nc d\ne f\n3 1 2\n输出：YES\n\n"
+            "当前问题输入：\n"
+            f"{example_input}\n\n"
+            "请将最终答案放在[answer]和[/answer]标签之间。"
+        )
+        return prompt
+    
+    @staticmethod
+    def extract_output(output):
+        # 支持多格式匹配（包含换行和大小写）
+        matches = re.findall(r'\[answer\s*\]\s*(.*?)\s*\[\s*/answer\s*\]', output, re.IGNORECASE | re.DOTALL)
+        if not matches:
+            return None
+        last_match = matches[-1].strip().upper()
+        return last_match if last_match in {"YES", "NO"} else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 验证前先进行基本清洗
+        cleaned_solution = str(solution).strip().upper()
+        return cleaned_solution == solve_handle_order(identity['names'], identity['p'])
+
+# 测试用例示例
+if __name__ == "__main__":
+    bootcamp = Cdesigntutorialmakeitnondeterministicbootcamp(solvable=True)
+    case = bootcamp.case_generator()
+    print("生成案例:", case)
+    print("问题描述:", bootcamp.prompt_func(case)[:200] + "...")
+    print("验证结果:", bootcamp._verify_correction("YES", case))