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183 lines
5.3 KiB
Python
Executable file
183 lines
5.3 KiB
Python
Executable file
"""#
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### 谜题描述
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Momiji has got a rooted tree, consisting of n nodes. The tree nodes are numbered by integers from 1 to n. The root has number 1. Momiji decided to play a game on this tree.
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The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by v) and removes all the subtree nodes with the root in node v from the tree. Node v gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.
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Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.
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Input
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The first line contains integer n (1 ≤ n ≤ 105) — the number of nodes in the tree. The next n - 1 lines contain the tree edges. The i-th line contains integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the numbers of the nodes that are connected by the i-th edge.
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It is guaranteed that the given graph is a tree.
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Output
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Print a single real number — the expectation of the number of steps in the described game.
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The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
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Examples
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Input
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2
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1 2
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Output
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1.50000000000000000000
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Input
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3
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1 2
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1 3
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Output
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2.00000000000000000000
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Note
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In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are:
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1 × (1 / 2) + 2 × (1 / 2) = 1.5
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In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are:
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1 × (1 / 3) + (1 + 1.5) × (2 / 3) = (1 / 3) + (5 / 3) = 2
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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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n = (int)(raw_input())
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ed = [[] for x in range(0, n)]
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vis = [0 for x in range(0, n)]
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for i in range(1, n):
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x, y = map(int, raw_input().split())
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x -= 1
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y -= 1
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ed[x].append(y)
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ed[y].append(x)
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ret = 0.0;
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d = [0 for x in range(0, n)]
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Q = [0]
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d[0] = 1
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vis[0] =1
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pos = 0
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while (pos < n):
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u = Q[pos]
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pos += 1
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ret += 1.0 / d[u]
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for v in ed[u]:
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if(vis[v] == 0):
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vis[v] = 1
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d[v] = d[u] + 1
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Q.append(v)
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print ret
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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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import random
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import re
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from collections import deque
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from bootcamp import Basebootcamp
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class Cgameontreebootcamp(Basebootcamp):
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def __init__(self, min_nodes=2, max_nodes=20):
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if min_nodes < 2 or max_nodes < min_nodes:
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raise ValueError("Node range must satisfy 2 ≤ min_nodes ≤ max_nodes")
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self.min_nodes = min_nodes
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self.max_nodes = max_nodes
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def case_generator(self):
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n = random.randint(self.min_nodes, self.max_nodes)
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edges = []
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parents = {}
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# Generate valid tree structure
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for node in range(2, n+1):
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parent = random.randint(1, node-1)
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parents[node] = parent
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edges.append((parent, node))
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# Build adjacency list
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adj = [[] for _ in range(n)]
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for a, b in edges:
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a_idx = a-1
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b_idx = b-1
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adj[a_idx].append(b_idx)
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adj[b_idx].append(a_idx)
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# BFS for depth calculation
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depths = [0]*n
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visited = [False]*n
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q = deque([0]) # Root node (1) has index 0
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visited[0] = True
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expectation = 0.0
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while q:
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u = q.popleft()
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expectation += 1.0 / (depths[u] + 1) # Depth starts from 0
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for v in adj[u]:
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if not visited[v]:
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visited[v] = True
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depths[v] = depths[u] + 1
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q.append(v)
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return {
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'n': n,
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'edges': edges,
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'expected': expectation,
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'_depth_info': depths # For debug purposes
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}
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@staticmethod
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def prompt_func(question_case):
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edges = '\n'.join(f"{a} {b}" for a, b in question_case['edges'])
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return f"""Given a rooted tree with {question_case['n']} nodes (root=1), calculate the expected number of steps to delete all nodes through random subtree removal.
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Input:
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{question_case['n']}
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{edges}
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Output requirements:
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- Compute expectation with 12+ decimal places
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- Enclose final answer in [answer][/answer]
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- Example: [answer]2.000000000000[/answer]"""
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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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if not matches:
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return None
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try:
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return float(matches[-1].strip())
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except (ValueError, TypeError):
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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if solution is None:
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return False
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expected = identity['expected']
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# Handle near-zero expected values
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if abs(expected) < 1e-12:
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return abs(solution) < 1e-6
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abs_error = abs(solution - expected)
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rel_error = abs_error / abs(expected)
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return abs_error <= 1e-6 or rel_error <= 1e-6
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