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223 lines
7.3 KiB
Python
Executable file
223 lines
7.3 KiB
Python
Executable file
"""#
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### 谜题描述
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Arkady bought an air ticket from a city A to a city C. Unfortunately, there are no direct flights, but there are a lot of flights from A to a city B, and from B to C.
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There are n flights from A to B, they depart at time moments a_1, a_2, a_3, ..., a_n and arrive at B t_a moments later.
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There are m flights from B to C, they depart at time moments b_1, b_2, b_3, ..., b_m and arrive at C t_b moments later.
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The connection time is negligible, so one can use the i-th flight from A to B and the j-th flight from B to C if and only if b_j ≥ a_i + t_a.
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You can cancel at most k flights. If you cancel a flight, Arkady can not use it.
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Arkady wants to be in C as early as possible, while you want him to be in C as late as possible. Find the earliest time Arkady can arrive at C, if you optimally cancel k flights. If you can cancel k or less flights in such a way that it is not possible to reach C at all, print -1.
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Input
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The first line contains five integers n, m, t_a, t_b and k (1 ≤ n, m ≤ 2 ⋅ 10^5, 1 ≤ k ≤ n + m, 1 ≤ t_a, t_b ≤ 10^9) — the number of flights from A to B, the number of flights from B to C, the flight time from A to B, the flight time from B to C and the number of flights you can cancel, respectively.
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The second line contains n distinct integers in increasing order a_1, a_2, a_3, ..., a_n (1 ≤ a_1 < a_2 < … < a_n ≤ 10^9) — the times the flights from A to B depart.
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The third line contains m distinct integers in increasing order b_1, b_2, b_3, ..., b_m (1 ≤ b_1 < b_2 < … < b_m ≤ 10^9) — the times the flights from B to C depart.
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Output
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If you can cancel k or less flights in such a way that it is not possible to reach C at all, print -1.
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Otherwise print the earliest time Arkady can arrive at C if you cancel k flights in such a way that maximizes this time.
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Examples
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Input
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4 5 1 1 2
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1 3 5 7
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1 2 3 9 10
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Output
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11
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Input
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2 2 4 4 2
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1 10
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10 20
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Output
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-1
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Input
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4 3 2 3 1
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1 999999998 999999999 1000000000
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3 4 1000000000
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Output
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1000000003
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Note
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Consider the first example. The flights from A to B depart at time moments 1, 3, 5, and 7 and arrive at B at time moments 2, 4, 6, 8, respectively. The flights from B to C depart at time moments 1, 2, 3, 9, and 10 and arrive at C at time moments 2, 3, 4, 10, 11, respectively. You can cancel at most two flights. The optimal solution is to cancel the first flight from A to B and the fourth flight from B to C. This way Arkady has to take the second flight from A to B, arrive at B at time moment 4, and take the last flight from B to C arriving at C at time moment 11.
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In the second example you can simply cancel all flights from A to B and you're done.
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In the third example you can cancel only one flight, and the optimal solution is to cancel the first flight from A to B. Note that there is still just enough time to catch the last flight from B to C.
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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 bisect, collections, heapq, sys
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n, m, ta, tb, k = map(int, sys.stdin.readline().rstrip().split(' '))
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a = map(int, sys.stdin.readline().rstrip().split(' '))
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b = map(int, sys.stdin.readline().rstrip().split(' '))
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if k >= n or k >= m:
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print(-1)
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sys.exit()
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ret = -1
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for i in range(k+1):
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val = a[i] + ta
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numDeleteB = k - i
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idx = bisect.bisect_left(b, val)
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if idx + numDeleteB >= m:
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print(-1)
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sys.exit()
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j = idx + numDeleteB
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ret = max(ret, b[j] + tb)
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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 bisect
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import random
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import re
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from bootcamp import Basebootcamp
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class Bbornthiswaybootcamp(Basebootcamp):
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def __init__(self, **params):
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default_params = {
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'n_min': 1,
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'n_max': 5,
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'm_min': 1,
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'm_max': 5,
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'ta_min': 1,
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'ta_max': 100,
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'tb_min': 1,
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'tb_max': 100,
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'k_min': 1,
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'k_max': None,
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}
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self.params = default_params.copy()
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self.params.update(params)
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def case_generator(self):
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n = random.randint(self.params['n_min'], self.params['n_max'])
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m = random.randint(self.params['m_min'], self.params['m_max'])
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ta = random.randint(self.params['ta_min'], self.params['ta_max'])
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tb = random.randint(self.params['tb_min'], self.params['tb_max'])
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max_k = n + m
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# Handle k constraints
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k_min = max(1, self.params['k_min'])
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k_min = min(k_min, max_k) # Ensure within valid range
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k_max_param = self.params['k_max']
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if k_max_param is None:
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effective_k_max = max_k
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else:
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effective_k_max = min(k_max_param, max_k)
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effective_k_max = max(k_min, effective_k_max)
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effective_k_max = min(effective_k_max, max_k) # Final validation
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k = random.randint(k_min, effective_k_max) if effective_k_max >= k_min else k_min
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# Generate strictly increasing flight times
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a = sorted(random.sample(range(1, 1000), n))
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b = sorted(random.sample(range(1, 1000), m))
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return {
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'n': n, 'm': m, 'ta': ta, 'tb': tb, 'k': k,
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'a': a, 'b': b
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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a_str = ', '.join(map(str, question_case['a']))
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b_str = ', '.join(map(str, question_case['b']))
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prompt = f"""航空调度谜题：
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- 从A到B有{question_case['n']}个航班（出发时间：{a_str}，飞行耗时{question_case['ta']}）
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- 从B到C有{question_case['m']}个航班（出发时间：{b_str}，飞行耗时{question_case['tb']}）
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- 你可取消最多{question_case['k']}个航班
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规则：
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1. 能转机的条件：B出发时间 ≥ A航班到达时间（A出发+{question_case['ta']}）
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2. 你的目标是让Arkady最晚到达C
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3. 若无法到达输出-1
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请计算最优解，并将最终答案用[answer]包裹。例如：[answer]11[/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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try:
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return int(matches[-1].strip())
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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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n, m = identity['n'], identity['m']
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ta, tb, k = identity['ta'], identity['tb'], identity['k']
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a, b = identity['a'], identity['b']
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# Edge case: impossible to connect
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if k >= n or k >= m:
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return solution == -1
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max_time = -1
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valid = False
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for x in range(k+1):
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if x >= n: continue
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# Find earliest possible B flight
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arrive_b = a[x] + ta
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idx = bisect.bisect_left(b, arrive_b)
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# Check remaining deletions in B flights
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remaining_deletions = k - x
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if idx + remaining_deletions >= m:
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continue
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# Calculate arrival time
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candidate = b[idx + remaining_deletions] + tb
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max_time = max(max_time, candidate)
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valid = True
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expected = max_time if valid else -1
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return solution == expected
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