You want to schedule a list of jobs in d
days. Jobs are dependent (i.e To work on the i-th
job, you have to finish all the jobs j
where 0 <= j < i
).
You have to finish at least one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the d
days. The difficulty of a day is the maximum difficulty of a job done in that day.
Given an array of integers jobDifficulty
and an integer d
. The difficulty of the i-th
job is jobDifficulty[i]
.
Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.
Example 1:
Input: jobDifficulty = [6,5,4,3,2,1], d = 2 Output: 7 Explanation: First day you can finish the first 5 jobs, total difficulty = 6. Second day you can finish the last job, total difficulty = 1. The difficulty of the schedule = 6 + 1 = 7
Example 2:
Input: jobDifficulty = [9,9,9], d = 4 Output: -1 Explanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.
Example 3:
Input: jobDifficulty = [1,1,1], d = 3 Output: 3 Explanation: The schedule is one job per day. total difficulty will be 3.
Example 4:
Input: jobDifficulty = [7,1,7,1,7,1], d = 3 Output: 15
Example 5:
Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6 Output: 843
Constraints:
1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10
题目的意思是给了一个数组,要求分成d组,并且满足每组最大值的和sum最小。
class Solution { public int minDifficulty(int[] jobDifficulty, int d) { int n = jobDifficulty.length; int[][] dp = new int[n + 1][d + 1]; if(d > n) return -1; for(int i = 0; i <= n; i++) Arrays.fill(dp[i], Integer.MAX_VALUE / 2); dp[0][0] = 0; for(int i = 1; i <= n; i++){ for(int k = 1; k <= d; k++){ int md = 0; for(int j = i - 1; j >= k - 1; j--){ md = Math.max(md, jobDifficulty[j]); dp[i][k] = Math.min(dp[i][k], dp[j][k - 1] + md); } } } return dp[n][d]; } }
https://www.bilibili.com/video/av85347641