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231 lines
7.9 KiB
Python
Executable file
231 lines
7.9 KiB
Python
Executable file
"""#
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### 谜题描述
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A top-secret military base under the command of Colonel Zuev is expecting an inspection from the Ministry of Defence. According to the charter, each top-secret military base must include a top-secret troop that should... well, we cannot tell you exactly what it should do, it is a top secret troop at the end. The problem is that Zuev's base is missing this top-secret troop for some reasons.
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The colonel decided to deal with the problem immediately and ordered to line up in a single line all n soldiers of the base entrusted to him. Zuev knows that the loquacity of the i-th soldier from the left is equal to qi. Zuev wants to form the top-secret troop using k leftmost soldiers in the line, thus he wants their total loquacity to be as small as possible (as the troop should remain top-secret). To achieve this, he is going to choose a pair of consecutive soldiers and swap them. He intends to do so no more than s times. Note that any soldier can be a participant of such swaps for any number of times. The problem turned out to be unusual, and colonel Zuev asked you to help.
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Determine, what is the minimum total loquacity of the first k soldiers in the line, that can be achieved by performing no more than s swaps of two consecutive soldiers.
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Input
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The first line of the input contains three positive integers n, k, s (1 ≤ k ≤ n ≤ 150, 1 ≤ s ≤ 109) — the number of soldiers in the line, the size of the top-secret troop to be formed and the maximum possible number of swap operations of the consecutive pair of soldiers, respectively.
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The second line of the input contains n integer qi (1 ≤ qi ≤ 1 000 000) — the values of loquacity of soldiers in order they follow in line from left to right.
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Output
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Print a single integer — the minimum possible total loquacity of the top-secret troop.
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Examples
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Input
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3 2 2
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2 4 1
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Output
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3
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Input
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5 4 2
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10 1 6 2 5
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Output
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18
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Input
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5 2 3
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3 1 4 2 5
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Output
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3
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Note
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In the first sample Colonel has to swap second and third soldiers, he doesn't really need the remaining swap. The resulting soldiers order is: (2, 1, 4). Minimum possible summary loquacity of the secret troop is 3. In the second sample Colonel will perform swaps in the following order:
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1. (10, 1, 6 — 2, 5)
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2. (10, 1, 2, 6 — 5)
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The resulting soldiers order is (10, 1, 2, 5, 6).
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Minimum possible summary loquacity is equal to 18.
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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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int n, k, s, ar[1 << 20];
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int dp[3][200][200 * 200];
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int main() {
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ios_base::sync_with_stdio(0);
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cin >> n >> k >> s;
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for (int i = 1; i <= n; i++) cin >> ar[i];
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if (s > n * n / 2 + 10) s = n * n / 2 + 10;
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for (int done = 0; done <= s; done++)
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for (int pref = 0; pref <= k; pref++) dp[0][pref][done] = 1e9;
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dp[0][0][0] = 0;
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for (int i = 1; i <= n; i++) {
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for (int j = 0; j <= s; j++)
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for (int q = 0; q <= k; q++) dp[i % 2][q][j] = 1e9;
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for (int pref = 0; pref < i && pref <= k; pref++)
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for (int done = 0; done <= s; done++) {
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dp[i % 2][pref][done] =
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min(dp[i % 2][pref][done], dp[1 - i % 2][pref][done]);
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int ndone = done + i - pref - 1;
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if (ndone <= s)
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dp[i % 2][pref + 1][ndone] = min(dp[i % 2][pref + 1][ndone],
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dp[1 - i % 2][pref][done] + ar[i]);
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}
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}
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int ans = 1e9;
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for (int i = 0; i <= s; i++) ans = min(ans, dp[n % 2][k][i]);
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cout << ans << endl;
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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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from bootcamp import Basebootcamp
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def calculate_min_loquacity(n, k, s, q):
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adjusted_s = min(s, (n*n)//2 + 10) # 严格模拟参考代码的调整逻辑
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INF = float('inf')
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# 初始化DP数组，使用滚动数组优化
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dp = [[[INF] * (adjusted_s + 1) for _ in range(k+1)] for __ in range(2)]
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dp[0][0][0] = 0 # 初始状态
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for i in range(1, n+1):
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current = i % 2
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prev = 1 - current
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# 重置当前层
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for j in range(k+1):
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for t in range(adjusted_s + 1):
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dp[current][j][t] = INF
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# 状态转移
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for pref in range(0, min(i-1, k)+1):
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for done in range(adjusted_s + 1):
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if dp[prev][pref][done] == INF:
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continue
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# 情况1：不选当前士兵
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if dp[current][pref][done] > dp[prev][pref][done]:
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dp[current][pref][done] = dp[prev][pref][done]
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# 情况2：选当前士兵
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new_pref = pref + 1
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if new_pref > k:
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continue
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swaps_needed = i - new_pref # 与参考代码完全一致的计算方式
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new_done = done + swaps_needed
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if new_done <= adjusted_s:
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new_value = dp[prev][pref][done] + q[i-1]
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if new_value < dp[current][new_pref][new_done]:
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dp[current][new_pref][new_done] = new_value
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# 寻找最终答案
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final_layer = n % 2
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return min(dp[final_layer][k][:adjusted_s+1])
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class Dtopsecrettaskbootcamp(Basebootcamp):
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def __init__(self, **params):
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self.params = params
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# 参数校验确保合法性
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self.params.setdefault('min_n', 3)
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self.params.setdefault('max_n', 150)
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self.params.setdefault('min_k', 1)
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self.params.setdefault('max_k', 150)
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self.params.setdefault('min_s', 1)
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self.params.setdefault('max_s', 10**9)
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self.params.setdefault('q_min', 1)
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self.params.setdefault('q_max', 10**6)
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# 确保参数间约束关系
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self.params['min_n'] = max(self.params['min_n'], 1)
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self.params['max_n'] = min(self.params['max_n'], 150)
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self.params['min_k'] = max(self.params['min_k'], 1)
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self.params['max_k'] = min(self.params['max_k'], self.params['max_n'])
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def case_generator(self):
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# 确保k <= n的约束
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n = random.randint(self.params['min_n'], self.params['max_n'])
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k = random.randint(
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self.params['min_k'],
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min(n, self.params['max_k'])
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)
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s = random.randint(self.params['min_s'], self.params['max_s'])
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# 生成loquacity值时保证至少k个非极大值
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q = [
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random.randint(self.params['q_min'], self.params['q_max'])
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for _ in range(n)
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]
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# 插入k个较小值以确保有解
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for _ in range(k):
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q[random.randint(0, n-1)] = random.randint(
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self.params['q_min'],
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self.params['q_max']//100
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)
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return {
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'n': n,
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'k': k,
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's': s,
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'q': q,
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'expected': calculate_min_loquacity(n, k, s, q)
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}
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@staticmethod
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def prompt_func(question_case):
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return f"""作为军事参谋，你需要解决秘密部队优化问题。请根据以下输入数据计算最小loquacity总和：
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输入格式：
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第一行：n k s
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第二行：q1 q2 ... qn
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当前输入：
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{question_case['n']} {question_case['k']} {question_case['s']}
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{' '.join(map(str, question_case['q']))}
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要求：
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1. 最多进行s次相邻交换
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2. 最终前k个士兵的loquacity总和最小
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3. 答案需用[answer]标签包裹，如：[answer]123[/answer]"""
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@staticmethod
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def extract_output(output):
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import re
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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 int(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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return solution == identity['expected']
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