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476 lines
11 KiB
Python
Executable file
476 lines
11 KiB
Python
Executable file
"""#
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### 谜题描述
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There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
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You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
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In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
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Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
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You have to compute m_i for all 1 ≤ i ≤ n.
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Input
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The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
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The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
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The description of each one of the k subsets follows, in the following format:
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The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
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The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
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It is guaranteed that:
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* The intersection of any three subsets is empty;
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* It's possible to make all lamps be simultaneously on using some operations.
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Output
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You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
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Examples
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Input
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7 3
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0011100
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3
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1 4 6
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3
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3 4 7
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2
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2 3
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Output
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1
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2
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3
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3
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3
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3
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3
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Input
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8 6
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00110011
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3
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1 3 8
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5
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1 2 5 6 7
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2
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6 8
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2
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3 5
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2
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4 7
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1
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2
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Output
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1
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1
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1
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1
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1
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1
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4
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4
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Input
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5 3
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00011
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3
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1 2 3
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1
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4
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3
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3 4 5
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Output
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1
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1
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1
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1
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1
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Input
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19 5
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1001001001100000110
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2
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2 3
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2
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5 6
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2
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8 9
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5
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12 13 14 15 16
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1
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19
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Output
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0
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1
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1
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1
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2
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2
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2
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3
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3
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3
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3
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4
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4
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4
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4
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4
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4
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4
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5
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Note
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In the first example:
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* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
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* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
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* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
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In the second example:
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* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
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* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
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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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n, k = [int(x) for x in input().split()]
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A = [+(c == '0') for c in input()]
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inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
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B = []
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for _ in range(k):
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m = inp[ii]; ii += 1
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B.append([a - 1 for a in inp[ii: ii + m]]); ii += m
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buckets = [[] for _ in range(n)]
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for j in range(k):
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for a in B[j]:
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buckets[a].append(j)
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#####################
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parent = list(range(2 * k))
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size = [1] * (2 * k)
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state = [0] * (2 * k)
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true = [1] * k + [0] * k
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cost = 0
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def find(a):
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acopy = a
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while a != parent[a]:
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a = parent[a]
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while acopy != a:
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parent[acopy], acopy = a, parent[acopy]
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return a
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def calcer(a):
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a = find(a)
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b = find((a + k) % (2 * k))
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if state[a] == 0:
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return min(true[a], true[b])
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elif state[a] == 1:
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return true[b]
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else:
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return true[a]
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def union(a, b):
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a, b = find(a), find(b)
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if a != b:
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if size[a] < size[b]:
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a, b = b, a
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parent[b] = a
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size[a] += size[b]
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state[a] |= state[b]
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true[a] += true[b]
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def on(a):
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global cost
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a = find(a)
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if state[a] == 0:
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b = find((a + k) % (2 * k))
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cost -= calcer(a)
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state[a] = 1
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state[b] = 2
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cost += calcer(a)
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def same(a, b):
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global cost
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a = find(a)
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b = find(b)
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if a != b:
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cost -= calcer(a)
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cost -= calcer(b)
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union(a, b)
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union((a + k) % (2 * k), (b + k) % (2 * k))
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cost += calcer(a)
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######
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out = []
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for i in range(n):
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bucket = buckets[i]
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if not bucket:
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pass
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elif len(bucket) == 1:
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j = bucket[0]
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if not A[i]:
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on(j)
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else:
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on(j + k)
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else:
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j1, j2 = bucket
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if not A[i]:
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same(j1, j2)
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else:
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same(j1, j2 + k)
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out.append(cost)
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print '\n'.join(str(x) for x in out)
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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 io import StringIO
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import sys
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from bootcamp import Basebootcamp
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class Cprefixenlightenmentbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.n_min = params.get('n_min', 5)
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self.n_max = params.get('n_max', 10)
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self.k_min = params.get('k_min', 1)
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self.k_max = params.get('k_max', 5)
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self.subset_size_min = params.get('subset_size_min', 1)
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self.subset_size_max = params.get('subset_size_max', 3)
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self.max_retries = params.get('max_retries', 1000)
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def case_generator(self):
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for _ in range(self.max_retries):
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n = random.randint(self.n_min, self.n_max)
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k = random.randint(self.k_min, self.k_max)
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count = [0] * n
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subsets = []
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valid = True
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for _ in range(k):
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c = random.randint(self.subset_size_min, min(self.subset_size_max, n))
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available = [x for x in range(n) if count[x] < 2]
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if len(available) < c:
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valid = False
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break
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selected = random.sample(available, c)
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subset = [x + 1 for x in selected]
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subsets.append(subset)
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for x in selected:
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count[x] += 1
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if not valid:
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continue
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x = [random.choice([0, 1]) for _ in range(k)]
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initial_state = []
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for i in range(n):
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sum_x = sum(x[j] for j in range(k) if (i + 1) in subsets[j])
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final_state = sum_x % 2
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initial_state.append(str(final_state ^ 1))
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initial_state_str = ''.join(initial_state)
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case = {
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'n': n,
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'k': k,
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'initial_state': initial_state_str,
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'subsets': subsets
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}
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return case
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raise ValueError("Failed to generate valid case after retries")
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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k = question_case['k']
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initial_state = question_case['initial_state']
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subsets = question_case['subsets']
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subset_str = '\n'.join([f"{len(subset)}\n{' '.join(map(str, subset))}" for subset in subsets])
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prompt = f"""You are solving a lamp switching puzzle. The task is to find the minimum operations required to turn on the first i lamps for all i from 1 to {n}.
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Problem Input:
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{n} {k}
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{initial_state}
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{subset_str}
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Rules:
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- Each switch toggles all lamps in its subset.
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- Any three subsets have empty intersection.
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- You need to output {n} lines where the i-th line is the minimum operations to turn on the first i lamps.
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Output your answer within [answer] and [/answer], each m_i on a separate line. Example:
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[answer]
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0
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1
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...
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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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solution = []
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for line in matches[-1].strip().split('\n'):
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line = line.strip()
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if line and line.isdigit():
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solution.append(int(line))
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elif line: # 包含非数字内容视为无效
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return None
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return solution # 长度验证在_verify_correction中进行
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@classmethod
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def _verify_correction(cls, solution, identity):
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if not solution or len(solution) != identity['n']:
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return False
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try:
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computed = cls.compute_solution(
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identity['n'], identity['k'],
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identity['initial_state'], identity['subsets']
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)
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return solution == computed
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except Exception as e:
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print(f"Verification error: {e}")
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return False
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@staticmethod
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def compute_solution(n, k, initial_state, subsets):
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input_lines = [f"{n} {k}", initial_state]
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for subset in subsets:
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input_lines.append(str(len(subset)))
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input_lines.append(' '.join(map(str, subset)))
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input_str = '\n'.join(input_lines)
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# 重新实现参考代码逻辑（Python3适配）
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from io import StringIO
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input_buf = StringIO(input_str)
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n_ref, k_ref = map(int, input_buf.readline().split())
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s_ref = input_buf.readline().strip()
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A_ref = [1 if c == '0' else 0 for c in s_ref]
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B_ref = []
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for _ in range(k_ref):
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c = int(input_buf.readline())
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elements = list(map(int, input_buf.readline().split()))
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B_ref.append([x - 1 for x in elements]) # 转换为0-based
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buckets = [[] for _ in range(n_ref)]
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for j in range(k_ref):
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for a in B_ref[j]:
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buckets[a].append(j)
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parent = list(range(2 * k_ref))
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size = [1] * (2 * k_ref)
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state = [0] * (2 * k_ref)
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true = [1] * k_ref + [0] * k_ref
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cost = 0
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def find(a):
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while parent[a] != a:
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parent[a] = parent[parent[a]] # 路径压缩
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a = parent[a]
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return a
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def calcer(a):
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a_root = find(a)
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b_root = find((a_root + k_ref) % (2 * k_ref))
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if state[a_root] == 0:
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return min(true[a_root], true[b_root])
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elif state[a_root] == 1:
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return true[b_root]
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else:
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return true[a_root]
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def union(a, b):
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nonlocal cost
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a_root = find(a)
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b_root = find(b)
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if a_root != b_root:
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# 按秩合并
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if size[a_root] < size[b_root]:
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a_root, b_root = b_root, a_root
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parent[b_root] = a_root
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size[a_root] += size[b_root]
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state[a_root] |= state[b_root]
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true[a_root] += true[b_root]
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def on(a):
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nonlocal cost
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a_root = find(a)
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if state[a_root] == 0:
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b_root = find((a_root + k_ref) % (2 * k_ref))
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cost -= calcer(a_root)
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state[a_root] = 1
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state[b_root] = 2
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cost += calcer(a_root)
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out = []
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for i in range(n_ref):
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bucket = buckets[i]
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if not bucket:
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pass
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elif len(bucket) == 1:
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j = bucket[0]
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if A_ref[i]:
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on(j)
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else:
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on(j + k_ref)
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else:
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j1, j2 = bucket
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if A_ref[i]:
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union(j1, j2)
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else:
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union(j1, j2 + k_ref)
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out.append(cost)
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return out
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