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internbootcamp/bootcamp/dcompletetripartite/dcompletetripartite.py

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+"""# 
+
+### 谜题描述
+You have a simple undirected graph consisting of n vertices and m edges. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.
+
+Let's make a definition.
+
+Let v_1 and v_2 be two some nonempty subsets of vertices that do not intersect. Let f(v_{1}, v_{2}) be true if and only if all the conditions are satisfied:
+
+  1. There are no edges with both endpoints in vertex set v_1. 
+  2. There are no edges with both endpoints in vertex set v_2. 
+  3. For every two vertices x and y such that x is in v_1 and y is in v_2, there is an edge between x and y. 
+
+
+
+Create three vertex sets (v_{1}, v_{2}, v_{3}) which satisfy the conditions below;
+
+  1. All vertex sets should not be empty. 
+  2. Each vertex should be assigned to only one vertex set. 
+  3. f(v_{1}, v_{2}), f(v_{2}, v_{3}), f(v_{3}, v_{1}) are all true. 
+
+
+
+Is it possible to create such three vertex sets? If it's possible, print matching vertex set for each vertex.
+
+Input
+
+The first line contains two integers n and m (3 ≤ n ≤ 10^{5}, 0 ≤ m ≤ min(3 ⋅ 10^{5}, (n(n-1))/(2))) — the number of vertices and edges in the graph.
+
+The i-th of the next m lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.
+
+Output
+
+If the answer exists, print n integers. i-th integer means the vertex set number (from 1 to 3) of i-th vertex. Otherwise, print -1.
+
+If there are multiple answers, print any.
+
+Examples
+
+Input
+
+
+6 11
+1 2
+1 3
+1 4
+1 5
+1 6
+2 4
+2 5
+2 6
+3 4
+3 5
+3 6
+
+
+Output
+
+
+1 2 2 3 3 3 
+
+Input
+
+
+4 6
+1 2
+1 3
+1 4
+2 3
+2 4
+3 4
+
+
+Output
+
+
+-1
+
+Note
+
+In the first example, if v_{1} = \{ 1 \}, v_{2} = \{ 2, 3 \}, and v_{3} = \{ 4, 5, 6 \} then vertex sets will satisfy all conditions. But you can assign vertices to vertex sets in a different way; Other answers like \"2 3 3 1 1 1\" will be accepted as well.
+
+<image>
+
+In the second example, it's impossible to make such vertex sets.
+
+Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
+```python
+import os
+import sys
+from atexit import register
+from io import BytesIO
+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')
+n,m = map(int,input().split(\" \"))
+connected = [[]for i in range(n+1)]
+edges = []
+for i in range(m):
+    a,b = map(int,input().split(\" \"))
+    connected[a].append(b)
+    connected[b].append(a)
+    edges.append((a,b))
+flag = True
+labels = [0]*(n+1)
+labels[1] = 1
+alln = set(range(1,n+1))
+if len(connected[1]) >0:
+    node2 = connected[1][0]
+else:
+    flag = False
+if flag:
+    s1 = alln-set(connected[1])
+    s2 = alln-set(connected[node2])
+    s3 = alln-s1-s2
+    n1,n2,n3 = len(s1),len(s2),len(s3)
+    if n1+n2+n3!=n or min(n1,n2,n3) == 0:
+        flag = False
+    for i in list(s1):
+        labels[i] = 1
+    for i in list(s2):
+        labels[i] = 2
+    for i in list(s3):
+        labels[i] = 3
+    cnts = [0]*3
+    for a,b in edges:
+        if labels[a] == labels[b]:
+            flag = False
+            break
+        cnts[labels[a]-1] += 1
+        cnts[labels[b]-1] += 1
+    if flag and cnts[0] == n1*(n2+n3) and cnts[1] == n2*(n1+n3) and cnts[2] == n3*(n1+n2):
+        out = []
+        for i in range(1,n+1):
+            out.append(str(labels[i]))
+        print \" \".join(out)
+    else:
+        print -1
+else:
+    print -1
+```
+
+
+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
+from bootcamp import Basebootcamp
+import re
+import random
+
+class Dcompletetripartitebootcamp(Basebootcamp):
+    def __init__(self):
+        super().__init__()
+    
+    def case_generator(self):
+        # 生成符合题目要求的n范围 (3 ≤ n ≤ 1e5)
+        n = random.randint(3, 20)  # 示例使用较小范围便于测试，实际可调至1e5
+        
+        # 确保三个子集都至少有一个顶点
+        sizes = [1, 1, 1]
+        remaining = n - 3
+        for _ in range(remaining):
+            sizes[random.randint(0, 2)] += 1
+        
+        # 随机分配顶点到三个子集
+        vertices = list(range(1, n+1))
+        random.shuffle(vertices)
+        
+        v1 = sorted(vertices[:sizes[0]])
+        v2 = sorted(vertices[sizes[0]:sizes[0]+sizes[1]])
+        v3 = sorted(vertices[sizes[0]+sizes[1]:])
+        
+        # 计算跨子集边的总数（数学公式直接计算）
+        m = sizes[0]*sizes[1] + sizes[1]*sizes[2] + sizes[2]*sizes[0]
+        
+        # 生成边集合（仅记录跨子集的边）
+        edges = []
+        # V1-V2边
+        for a in v1:
+            edges.extend((min(a,b), max(a,b)) for b in v2)
+        # V2-V3边
+        for b in v2:
+            edges.extend((min(b,c), max(b,c)) for c in v3)
+        # V3-V1边
+        for c in v3:
+            edges.extend((min(c,a), max(c,a)) for a in v1)
+        
+        # 去重并排序
+        edges = sorted(list(set(edges)))
+        
+        return {
+            "n": n,
+            "m": len(edges),
+            "edges": edges,
+            "expected_sets": {
+                1: v1,
+                2: v2,
+                3: v3
+            }
+        }
+    
+    @staticmethod
+    def prompt_func(question_case) -> str:
+        input_lines = [f"{question_case['n']} {question_case['m']}"]
+        input_lines.extend(f"{a} {b}" for a, b in question_case['edges'])
+        input_block = "\n".join(input_lines)
+        
+        prompt = f"""Given an undirected graph with {question_case['n']} vertices and {question_case['m']} edges, determine if the vertices can be partitioned into three non-empty subsets meeting these conditions:
+
+1. All vertices are in exactly one subset
+2. For each subset pair (V1,V2), (V2,V3), (V3,V1):
+   - No internal edges within either subset
+   - Complete bipartite connections between subsets
+
+Input:
+{input_block}
+
+Output format:
+If possible: {question_case['n']} space-separated integers (1/2/3)
+If impossible: -1
+
+Place your final answer between [answer] and [/answer]. Example:
+[answer]1 2 2 3 3 3[/answer]"""
+
+        return prompt
+    
+    @staticmethod
+    def extract_output(output):
+        matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
+        return matches[-1].strip() if matches else None
+    
+    @classmethod
+    def _verify_correction(cls, solution, identity):
+        # 快速拒绝错误格式
+        if not solution or solution.strip() == "-1":
+            return False
+        
+        try:
+            labels = list(map(int, solution.split()))
+        except ValueError:
+            return False
+        
+        # 验证基础条件
+        if len(labels) != identity['n']:
+            return False
+        if not all(l in {1,2,3} for l in labels):
+            return False
+        
+        # 构建分组映射
+        groups = {1: set(), 2: set(), 3: set()}
+        for v, l in enumerate(labels, 1):
+            groups[l].add(v)
+        
+        # 检查非空子集
+        if any(len(groups[i]) == 0 for i in [1,2,3]):
+            return False
+        
+        # 验证边约束
+        edge_set = set(map(tuple, identity['edges']))
+        
+        # 验证每个子集的内部无连接
+        for group in groups.values():
+            for a in group:
+                for b in group:
+                    if a < b and (a, b) in edge_set:
+                        return False
+        
+        # 验证跨子集的完全连接
+        expected_pairs = [
+            (groups[1], groups[2]),
+            (groups[2], groups[3]),
+            (groups[3], groups[1])
+        ]
+        for s1, s2 in expected_pairs:
+            for a in s1:
+                for b in s2:
+                    if a > b:
+                        a, b = b, a
+                    if (a, b) not in edge_set:
+                        return False
+        
+        return True