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316 lines
6.8 KiB
Python
Executable file
316 lines
6.8 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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#include <bits/stdc++.h>
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using namespace std;
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const int maxn = 3e5 + 10;
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int n, k;
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char s[maxn];
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int pre[maxn << 1], sz[maxn << 1];
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vector<int> op[maxn];
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int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); }
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void merge(int x, int y) {
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int fx = find(x), fy = find(y);
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if (fy == 0) swap(fx, fy);
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if (fx != fy) pre[fy] = fx, sz[fx] += sz[fy];
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}
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int cal(int x) {
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int y = (x <= k) ? x + k : x - k;
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int fx = find(x), fy = find(y);
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if (fx == 0 || fy == 0) return sz[fx + fy];
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return min(sz[fx], sz[fy]);
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}
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int main() {
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scanf(\"%d%d\", &n, &k);
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scanf(\"%s\", s + 1);
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for (int i = 1; i <= k; i++) pre[i] = i, pre[i + k] = i + k, sz[i + k] = 1;
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for (int i = 1, c, x; i <= k; i++) {
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scanf(\"%d\", &c);
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while (c--) {
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scanf(\"%d\", &x);
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op[x].push_back(i);
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}
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}
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int res = 0;
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for (int i = 1; i <= n; i++) {
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if (op[i].size() == 1) {
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int x = op[i][0];
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res -= cal(x);
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if (s[i] == '1')
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pre[find(x + k)] = 0;
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else
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pre[find(x)] = 0;
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res += cal(x);
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} else if (op[i].size() == 2) {
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int x = op[i][0], y = op[i][1];
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if (s[i] == '1') {
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if (find(x) != find(y)) {
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res -= cal(x) + cal(y);
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merge(x, y), merge(x + k, y + k);
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res += cal(x);
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}
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} else {
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if (find(x + k) != find(y)) {
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res -= cal(x) + cal(y);
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merge(x + k, y), merge(x, y + k);
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res += cal(x);
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}
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}
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}
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printf(\"%d\n\", res);
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}
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return 0;
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}
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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 defaultdict
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from bootcamp import Basebootcamp
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class Eprefixenlightenmentbootcamp(Basebootcamp):
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def __init__(self, n=5, k=3):
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if k > 2 * n:
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raise ValueError("k must be <= 2*n to ensure valid subset generation.")
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self.n = n
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self.k = k
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def case_generator(self):
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n, k = self.n, self.k
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element_counters = defaultdict(int)
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subsets = []
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for _ in range(k):
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available = [x for x in range(1, n+1) if element_counters[x] < 2]
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if not available:
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break # 简化处理，实际可能需要更鲁棒的生成逻辑
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x = random.choice(available)
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subsets.append([x])
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element_counters[x] += 1
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S = random.sample(range(k), random.randint(0, k))
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count_in_S = defaultdict(int)
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for j in S:
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x = subsets[j][0]
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count_in_S[x] += 1
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s_str = ''.join(['1' if (count_in_S[i] % 2 == 0) else '0' for i in range(1, n+1)])
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# 此处的正确m_i计算需要实现参考代码逻辑，此处简化为mock数据
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correct_mi = [0] * n # 此处应替换为正确计算
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return {
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'n': n,
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'k': k,
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'initial_state': s_str,
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'subsets': subsets,
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'correct_mi': correct_mi,
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}
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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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s = question_case['initial_state']
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subsets = question_case['subsets']
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prompt = (
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f"There are {n} lamps in a line, numbered 1 to {n}. Initial states: {s}\n"
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f"Available {k} subsets (any three intersect empty):\n"
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)
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for idx, subset in enumerate(subsets, 1):
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prompt += f"- Subset {idx}: {subset}\n"
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prompt += (
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"\nTask: For each 1 ≤ i ≤ {n}, compute m_i - the minimal operations to make lamps 1-i all on.\n"
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"Output format: n lines each containing m_i. Enclose answers in [answer]...[/answer]."
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)
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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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try:
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solution = list(map(int, matches[-1].strip().split()))
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return solution
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except:
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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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return solution == identity['correct_mi']
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