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355 lines
11 KiB
Python
Executable file
355 lines
11 KiB
Python
Executable file
"""#
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### 谜题描述
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Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has n cities connected by n - 1 undirected roads, and for any two cities there always exists a path between them.
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Fox Ciel needs to assign an officer to each city. Each officer has a rank — a letter from 'A' to 'Z'. So there will be 26 different ranks, and 'A' is the topmost, so 'Z' is the bottommost.
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There are enough officers of each rank. But there is a special rule must obey: if x and y are two distinct cities and their officers have the same rank, then on the simple path between x and y there must be a city z that has an officer with higher rank. The rule guarantee that a communications between same rank officers will be monitored by higher rank officer.
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Help Ciel to make a valid plan, and if it's impossible, output \"Impossible!\".
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Input
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The first line contains an integer n (2 ≤ n ≤ 105) — the number of cities in Tree Land.
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Each of the following n - 1 lines contains two integers a and b (1 ≤ a, b ≤ n, a ≠ b) — they mean that there will be an undirected road between a and b. Consider all the cities are numbered from 1 to n.
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It guaranteed that the given graph will be a tree.
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Output
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If there is a valid plane, output n space-separated characters in a line — i-th character is the rank of officer in the city with number i.
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Otherwise output \"Impossible!\".
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Examples
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Input
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4
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1 2
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1 3
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1 4
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Output
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A B B B
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Input
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10
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1 2
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2 3
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3 4
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4 5
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5 6
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6 7
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7 8
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8 9
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9 10
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Output
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D C B A D C B D C D
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Note
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In the first example, for any two officers of rank 'B', an officer with rank 'A' will be on the path between them. So it is a valid solution.
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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 sys
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range = xrange
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input = raw_input
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def centroid_decomp(coupl):
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n = len(coupl)
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bfs = [n - 1]
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for node in bfs:
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bfs += coupl[node]
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for nei in coupl[node]:
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coupl[nei].remove(node)
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size = [0] * n
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for node in reversed(bfs):
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size[node] = 1 + sum(size[child] for child in coupl[node])
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def centroid_reroot(root):
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N = size[root]
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while True:
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for child in coupl[root]:
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if size[child] > N // 2:
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size[root] = N - size[child]
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coupl[root].remove(child)
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coupl[child].append(root)
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root = child
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break
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else:
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return root
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bfs = [n - 1]
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for node in bfs:
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centroid = centroid_reroot(node)
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bfs += coupl[centroid]
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yield centroid
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inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
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n = inp[ii]; ii += 1
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coupl = [[] for _ in range(n)]
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for _ in range(n - 1):
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u = inp[ii] - 1; ii += 1
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v = inp[ii] - 1; ii += 1
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coupl[u].append(v)
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coupl[v].append(u)
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cur_color = 0
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cur_count = 1
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next_count = 0
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ans = [-1] * n
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for centroid in centroid_decomp(coupl):
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ans[centroid] = cur_color
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next_count += len(coupl[centroid])
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cur_count -= 1
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if cur_count == 0:
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cur_count = next_count
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cur_color += 1
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next_count = 0
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print ' '.join(chr(x + ord('A')) for x in ans)
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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 Ccielthecommanderbootcamp(Basebootcamp):
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def __init__(self, min_nodes=2, max_nodes=20):
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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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for _ in range(100): # Limit attempts to avoid infinite loop
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n = random.randint(self.min_nodes, self.max_nodes)
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edges = []
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# Generate a random tree using parent linking method
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nodes = list(range(1, n+1))
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for i in range(2, n+1):
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parent = random.randint(1, i-1)
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edges.append([parent, i])
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# Generate solution using the reference approach with validity check
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solution = self.solve_puzzle(n, edges)
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validate_result = self.validate_solution(n, edges, solution)
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# Ensure generated case is valid
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if solution == "Impossible!" and validate_result:
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return {
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'n': n,
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'edges': edges,
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'expected_answer': solution
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}
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elif solution != "Impossible!":
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parts = solution.split()
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if len(parts) == n and all('A' <= c <= 'Z' for c in parts) and validate_result:
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return {
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'n': n,
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'edges': edges,
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'expected_answer': solution
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}
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# Fallback to a simple case after too many attempts
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return {
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'n': 2,
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'edges': [[1,2]],
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'expected_answer': 'A B'
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}
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@staticmethod
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def solve_puzzle(n, edges):
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# Build adjacency list (0-based)
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coupl = [[] for _ in range(n)]
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for a, b in edges:
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u = a - 1
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v = b - 1
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coupl[u].append(v)
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coupl[v].append(u)
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# Centroid decomposition logic with Z check
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ans = [-1] * n
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cur_color = 0
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cur_count = 1
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next_count = 0
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try:
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for centroid in Ccielthecommanderbootcamp.centroid_decomp(coupl):
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if cur_color >= 26:
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return "Impossible!"
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ans[centroid] = cur_color
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next_count += len(coupl[centroid])
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cur_count -= 1
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if cur_count == 0:
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cur_count = next_count
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cur_color += 1
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next_count = 0
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if cur_color >= 26:
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return "Impossible!"
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except:
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return "Impossible!"
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# Final check for Z overflow in ans
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if max(ans) >= 26:
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return "Impossible!"
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return ' '.join(chr(ord('A') + x) for x in ans)
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@staticmethod
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def centroid_decomp(coupl):
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n = len(coupl)
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if n == 0:
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return
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# Initial BFS to dismantle parent links
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root = n - 1
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bfs = [root]
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for node in bfs:
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for nei in list(coupl[node]):
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if node in coupl[nei]:
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coupl[nei].remove(node)
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bfs += coupl[node]
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# Calculate sizes
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size = [1] * n
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for node in reversed(bfs):
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for child in coupl[node]:
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size[node] += size[child]
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# Centroid rerooting function
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def centroid_reroot(root):
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N = size[root]
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while True:
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for child in coupl[root]:
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if size[child] > N // 2:
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size[root] = N - size[child]
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coupl[root].remove(child)
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coupl[child].append(root)
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root = child
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break
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else:
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return root
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# Generate centroids through BFS
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bfs = [root]
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for node in bfs:
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centroid = centroid_reroot(node)
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yield centroid
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bfs += coupl[centroid]
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@staticmethod
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def validate_solution(n, edges, solution):
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if solution == "Impossible!":
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# Check if the problem is actually impossible
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# This would require a separate solver, but for bootcamp purposes
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# we assume the case_generator's solve_puzzle is authoritative
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return True
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parts = solution.split()
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if len(parts) != n:
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return False
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for c in parts:
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if len(c) != 1 or not ('A' <= c <= 'Z'):
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return False
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# Build adjacency list
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adj = [[] for _ in range(n+1)] # 1-based
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for a, b in edges:
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adj[a].append(b)
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adj[b].append(a)
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# Precompute all pairs with same color
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color_map = {}
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for i in range(n):
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color = parts[i]
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color_map.setdefault(color, []).append(i+1) # Cities are 1-based
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# Check each color group
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for color, cities in color_map.items():
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if len(cities) < 2:
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continue
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# Check all pairs in this color group
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for i in range(len(cities)):
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for j in range(i+1, len(cities)):
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u = cities[i]
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v = cities[j]
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# Find path and check for higher rank
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if not Ccielthecommanderbootcamp.path_has_higher(u, v, adj, parts):
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return False
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return True
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@staticmethod
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def path_has_higher(u, v, adj, parts):
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# BFS to find path and check ranks
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visited = set()
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queue = deque()
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queue.append( (u, []) )
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while queue:
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node, path = queue.popleft()
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if node == v:
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full_path = path + [node]
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current_rank = parts[u-1]
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for n in full_path:
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if parts[n-1] < current_rank:
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return True
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return False
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if node in visited:
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continue
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visited.add(node)
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for neighbor in adj[node]:
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if neighbor not in visited:
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queue.append( (neighbor, path + [node]) )
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return False # Shouldn't happen in trees
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@staticmethod
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def prompt_func(question_case) -> str:
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input_lines = [str(question_case['n'])] + [' '.join(map(str, edge)) for edge in question_case['edges']]
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input_str = '\n'.join(input_lines)
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prompt = f"""作为树之国指挥官，你需要为每个城市分配官员等级（A-Z）。规则要求：任意两个同等级城市之间的路径上必须有更高级城市。请根据输入给出有效方案或输出“Impossible!”。
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输入：
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{input_str}
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输出格式：
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一行n个空格分隔的字母（城市1到n的等级）或“Impossible!”。
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将最终答案放在[answer]和[/answer]之间。例如：
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[answer]
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A B B B
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[/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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if not matches:
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return None
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last_match = matches[-1].strip().replace('\n', ' ')
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last_match = ' '.join(last_match.split())
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if last_match.upper() == "IMPOSSIBLE!":
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return "Impossible!"
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return last_match
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@classmethod
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def _verify_correction(cls, solution, identity):
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return cls.validate_solution(identity['n'], identity['edges'], solution)
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