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284 lines
7.9 KiB
Python
Executable file
284 lines
7.9 KiB
Python
Executable file
"""#
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### 谜题描述
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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.
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Let's make a definition.
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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:
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1. There are no edges with both endpoints in vertex set v_1.
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2. There are no edges with both endpoints in vertex set v_2.
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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.
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Create three vertex sets (v_{1}, v_{2}, v_{3}) which satisfy the conditions below;
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1. All vertex sets should not be empty.
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2. Each vertex should be assigned to only one vertex set.
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3. f(v_{1}, v_{2}), f(v_{2}, v_{3}), f(v_{3}, v_{1}) are all true.
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Is it possible to create such three vertex sets? If it's possible, print matching vertex set for each vertex.
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Input
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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.
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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.
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Output
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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.
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If there are multiple answers, print any.
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Examples
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Input
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6 11
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1 2
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1 3
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1 4
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1 5
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1 6
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2 4
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2 5
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2 6
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3 4
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3 5
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3 6
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Output
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1 2 2 3 3 3
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Input
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4 6
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1 2
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1 3
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1 4
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2 3
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2 4
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3 4
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Output
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-1
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Note
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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.
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<image>
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In the second example, it's impossible to make such vertex sets.
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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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import os
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import sys
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from atexit import register
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from io import BytesIO
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sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
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sys.stdout = BytesIO()
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register(lambda: os.write(1, sys.stdout.getvalue()))
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input = lambda: sys.stdin.readline().rstrip('\r\n')
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n,m = map(int,input().split(\" \"))
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connected = [[]for i in range(n+1)]
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edges = []
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for i in range(m):
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a,b = map(int,input().split(\" \"))
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connected[a].append(b)
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connected[b].append(a)
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edges.append((a,b))
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flag = True
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labels = [0]*(n+1)
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labels[1] = 1
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alln = set(range(1,n+1))
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if len(connected[1]) >0:
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node2 = connected[1][0]
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else:
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flag = False
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if flag:
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s1 = alln-set(connected[1])
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s2 = alln-set(connected[node2])
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s3 = alln-s1-s2
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n1,n2,n3 = len(s1),len(s2),len(s3)
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if n1+n2+n3!=n or min(n1,n2,n3) == 0:
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flag = False
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for i in list(s1):
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labels[i] = 1
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for i in list(s2):
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labels[i] = 2
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for i in list(s3):
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labels[i] = 3
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cnts = [0]*3
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for a,b in edges:
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if labels[a] == labels[b]:
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flag = False
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break
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cnts[labels[a]-1] += 1
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cnts[labels[b]-1] += 1
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if flag and cnts[0] == n1*(n2+n3) and cnts[1] == n2*(n1+n3) and cnts[2] == n3*(n1+n2):
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out = []
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for i in range(1,n+1):
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out.append(str(labels[i]))
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print \" \".join(out)
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else:
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print -1
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else:
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print -1
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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 re
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import random
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class Dcompletetripartitebootcamp(Basebootcamp):
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def __init__(self):
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super().__init__()
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def case_generator(self):
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# 生成符合题目要求的n范围 (3 ≤ n ≤ 1e5)
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n = random.randint(3, 20) # 示例使用较小范围便于测试，实际可调至1e5
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# 确保三个子集都至少有一个顶点
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sizes = [1, 1, 1]
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remaining = n - 3
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for _ in range(remaining):
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sizes[random.randint(0, 2)] += 1
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# 随机分配顶点到三个子集
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vertices = list(range(1, n+1))
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random.shuffle(vertices)
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v1 = sorted(vertices[:sizes[0]])
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v2 = sorted(vertices[sizes[0]:sizes[0]+sizes[1]])
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v3 = sorted(vertices[sizes[0]+sizes[1]:])
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# 计算跨子集边的总数（数学公式直接计算）
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m = sizes[0]*sizes[1] + sizes[1]*sizes[2] + sizes[2]*sizes[0]
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# 生成边集合（仅记录跨子集的边）
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edges = []
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# V1-V2边
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for a in v1:
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edges.extend((min(a,b), max(a,b)) for b in v2)
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# V2-V3边
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for b in v2:
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edges.extend((min(b,c), max(b,c)) for c in v3)
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# V3-V1边
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for c in v3:
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edges.extend((min(c,a), max(c,a)) for a in v1)
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# 去重并排序
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edges = sorted(list(set(edges)))
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return {
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"n": n,
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"m": len(edges),
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"edges": edges,
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"expected_sets": {
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1: v1,
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2: v2,
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3: v3
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}
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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input_lines = [f"{question_case['n']} {question_case['m']}"]
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input_lines.extend(f"{a} {b}" for a, b in question_case['edges'])
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input_block = "\n".join(input_lines)
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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:
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1. All vertices are in exactly one subset
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2. For each subset pair (V1,V2), (V2,V3), (V3,V1):
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- No internal edges within either subset
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- Complete bipartite connections between subsets
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Input:
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{input_block}
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Output format:
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If possible: {question_case['n']} space-separated integers (1/2/3)
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If impossible: -1
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Place your final answer between [answer] and [/answer]. Example:
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[answer]1 2 2 3 3 3[/answer]"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL)
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return matches[-1].strip() if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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# 快速拒绝错误格式
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if not solution or solution.strip() == "-1":
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return False
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try:
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labels = list(map(int, solution.split()))
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except ValueError:
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return False
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# 验证基础条件
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if len(labels) != identity['n']:
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return False
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if not all(l in {1,2,3} for l in labels):
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return False
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# 构建分组映射
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groups = {1: set(), 2: set(), 3: set()}
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for v, l in enumerate(labels, 1):
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groups[l].add(v)
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# 检查非空子集
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if any(len(groups[i]) == 0 for i in [1,2,3]):
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return False
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# 验证边约束
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edge_set = set(map(tuple, identity['edges']))
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# 验证每个子集的内部无连接
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for group in groups.values():
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for a in group:
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for b in group:
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if a < b and (a, b) in edge_set:
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return False
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# 验证跨子集的完全连接
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expected_pairs = [
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(groups[1], groups[2]),
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(groups[2], groups[3]),
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(groups[3], groups[1])
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]
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for s1, s2 in expected_pairs:
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for a in s1:
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for b in s2:
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if a > b:
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a, b = b, a
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if (a, b) not in edge_set:
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return False
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return True
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