1335. Minimum Difficulty of a Job Schedule

Source: https://leetcode.com/contest/weekly-contest-173/problems/minimum-difficulty-of-a-job-schedule/

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

思路：DP

$dp[i][j]$dp[i][j]表示$j$j天安排$[0...i]$[0...i]的jobs，最小的difficulty。需要满足$i+1\geq j$i+1≥j.

当$j\geq 2$j≥2，$i\geq j-1$i≥j−1时，
$dp[i][j]=\min\limits_{j-2\leq k<i} (\max(jobDifficulty[k+1...i])+dp[k][j-1])$dp[i][j]=j−2≤k<imin​(max(jobDifficulty[k+1...i])+dp[k][j−1])
所以循环是$1\leq i< n$1≤i<n, $2\leq j\leq \min(i+1,d)$2≤j≤min(i+1,d).

对于$j==1$j==1时的所有$0\leq i< n$0≤i<n，就是取数组$[0...i]$[0...i]最大值即可。

class Solution {
public:
    int minDifficulty(vector<int>& jobDifficulty, int d) {
        int n=jobDifficulty.size();
        int INF=0x3f3f3f3f;
        vector<vector<int> > dp(n, vector<int>(d+1, INF));
        int maxdif=jobDifficulty[0];
        for(int i=0;i<n;i++){
            maxdif=max(maxdif,jobDifficulty[i]);
            dp[i][1]=maxdif;
        }
        for(int i=1;i<n;i++){
            for(int j=2;j<=i+1&&j<=d;j++){
                int maxdiff=jobDifficulty[i];
                for(int k=i-1;k>=j-2;k--){
                    maxdiff=max(jobDifficulty[k+1],maxdiff);
                    dp[i][j]=min(dp[i][j],maxdiff+dp[k][j-1]);
                }
            }
        }
        if(dp[n-1][d]==INF){
            dp[n-1][d]=-1;
        }
        return dp[n-1][d];
    }
};