{ "data": { "question": { "questionId": "3833", "questionFrontendId": "3538", "categoryTitle": "Algorithms", "boundTopicId": 3667711, "title": "Merge Operations for Minimum Travel Time", "titleSlug": "merge-operations-for-minimum-travel-time", "content": "
You are given a straight road of length l km, an integer n, an integer k, and two integer arrays, position and time, each of length n.
The array position lists the positions (in km) of signs in strictly increasing order (with position[0] = 0 and position[n - 1] = l).
Each time[i] represents the time (in minutes) required to travel 1 km between position[i] and position[i + 1].
You must perform exactly k merge operations. In one merge, you can choose any two adjacent signs at indices i and i + 1 (with i > 0 and i + 1 < n) and:
i + 1 so that its time becomes time[i] + time[i + 1].i.Return the minimum total travel time (in minutes) to travel from 0 to l after exactly k merges.
\n
Example 1:
\n\nInput: l = 10, n = 4, k = 1, position = [0,3,8,10], time = [5,8,3,6]
\n\nOutput: 62
\n\nExplanation:
\n\nMerge the signs at indices 1 and 2. Remove the sign at index 1, and change the time at index 2 to 8 + 3 = 11.
position array: [0, 8, 10]time array: [5, 11, 6]| Segment | \n\t\t\t\tDistance (km) | \n\t\t\t\tTime per km (min) | \n\t\t\t\tSegment Travel Time (min) | \n\t\t\t
|---|---|---|---|
| 0 → 8 | \n\t\t\t\t8 | \n\t\t\t\t5 | \n\t\t\t\t8 × 5 = 40 | \n\t\t\t
| 8 → 10 | \n\t\t\t\t2 | \n\t\t\t\t11 | \n\t\t\t\t2 × 11 = 22 | \n\t\t\t
40 + 22 = 62, which is the minimum possible time after exactly 1 merge.Example 2:
\n\nInput: l = 5, n = 5, k = 1, position = [0,1,2,3,5], time = [8,3,9,3,3]
\n\nOutput: 34
\n\nExplanation:
\n\n3 + 9 = 12.position array: [0, 2, 3, 5]time array: [8, 12, 3, 3]| Segment | \n\t\t\t\tDistance (km) | \n\t\t\t\tTime per km (min) | \n\t\t\t\tSegment Travel Time (min) | \n\t\t\t
|---|---|---|---|
| 0 → 2 | \n\t\t\t\t2 | \n\t\t\t\t8 | \n\t\t\t\t2 × 8 = 16 | \n\t\t\t
| 2 → 3 | \n\t\t\t\t1 | \n\t\t\t\t12 | \n\t\t\t\t1 × 12 = 12 | \n\t\t\t
| 3 → 5 | \n\t\t\t\t2 | \n\t\t\t\t3 | \n\t\t\t\t2 × 3 = 6 | \n\t\t\t
16 + 12 + 6 = 34, which is the minimum possible time after exactly 1 merge.\n
Constraints:
\n\n1 <= l <= 1052 <= n <= min(l + 1, 50)0 <= k <= min(n - 2, 10)position.length == nposition[0] = 0 and position[n - 1] = lposition is sorted in strictly increasing order.time.length == n1 <= time[i] <= 1001 <= sum(time) <= 100给你一个长度为 l 公里的直路,一个整数 n,一个整数 k 和 两个 长度为 n 的整数数组 position 和 time 。
数组 position 列出了路标的位置(单位:公里),并且是 严格 升序排列的(其中 position[0] = 0 且 position[n - 1] = l)。
每个 time[i] 表示从 position[i] 到 position[i + 1] 之间行驶 1 公里所需的时间(单位:分钟)。
你 必须 执行 恰好 k 次合并操作。在一次合并中,你可以选择两个相邻的路标,下标为 i 和 i + 1(其中 i > 0 且 i + 1 < n),并且:
i + 1 的路标,使其时间变为 time[i] + time[i + 1]。i 的路标。返回经过 恰好 k 次合并后从 0 到 l 的 最小总旅行时间(单位:分钟)。
\n\n
示例 1:
\n\n输入: l = 10, n = 4, k = 1, position = [0,3,8,10], time = [5,8,3,6]
\n\n输出: 62
\n\n解释:
\n\n合并下标为 1 和 2 的路标。删除下标为 1 的路标,并将下标为 2 的路标的时间更新为 8 + 3 = 11。
position 数组:[0, 8, 10]time 数组:[5, 11, 6]| 路段 | \n\t\t\t\t距离(公里) | \n\t\t\t\t每公里时间(分钟) | \n\t\t\t\t路段旅行时间(分钟) | \n\t\t\t
|---|---|---|---|
| 0 → 8 | \n\t\t\t\t8 | \n\t\t\t\t5 | \n\t\t\t\t8 × 5 = 40 | \n\t\t\t
| 8 → 10 | \n\t\t\t\t2 | \n\t\t\t\t11 | \n\t\t\t\t2 × 11 = 22 | \n\t\t\t
40 + 22 = 62 ,这是执行 1 次合并后的最小时间。示例 2:
\n\n输入: l = 5, n = 5, k = 1, position = [0,1,2,3,5], time = [8,3,9,3,3]
\n\n输出: 34
\n\n解释:
\n\n3 + 9 = 12。position 数组:[0, 2, 3, 5]time 数组:[8, 12, 3, 3]| 路段 | \n\t\t\t\t距离(公里) | \n\t\t\t\t每公里时间(分钟) | \n\t\t\t\t路段旅行时间(分钟) | \n\t\t\t
|---|---|---|---|
| 0 → 2 | \n\t\t\t\t2 | \n\t\t\t\t8 | \n\t\t\t\t2 × 8 = 16 | \n\t\t\t
| 2 → 3 | \n\t\t\t\t1 | \n\t\t\t\t12 | \n\t\t\t\t1 × 12 = 12 | \n\t\t\t
| 3 → 5 | \n\t\t\t\t2 | \n\t\t\t\t3 | \n\t\t\t\t2 × 3 = 6 | \n\t\t\t
16 + 12 + 6 = 34 ,这是执行 1 次合并后的最小时间。\n\n
提示:
\n\n1 <= l <= 1052 <= n <= min(l + 1, 50)0 <= k <= min(n - 2, 10)position.length == nposition[0] = 0 和 position[n - 1] = lposition 是严格升序排列的。time.length == n1 <= time[i] <= 1001 <= sum(time) <= 100k merges, you’ll have n-k signs left.",
"Define DP[i][j][s] as the minimum travel time for positions 0..i when i is kept, j deletions are done overall, and s consecutive deletions occurred immediately before i.",
"Update the DP by either merging (increment s and j) or not merging (reset s) and adding the appropriate travel time."
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