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1335. 工作计划的最低难度

题目描述

你需要制定一份 d 天的工作计划表。工作之间存在依赖,要想执行第 i 项工作,你必须完成全部 j 项工作( 0 <= j < i)。

你每天 至少 需要完成一项任务。工作计划的总难度是这 d 天每一天的难度之和,而一天的工作难度是当天应该完成工作的最大难度。

给你一个整数数组 jobDifficulty 和一个整数 d,分别代表工作难度和需要计划的天数。第 i 项工作的难度是 jobDifficulty[i]

返回整个工作计划的 最小难度 。如果无法制定工作计划,则返回 -1 

 

示例 1:

输入:jobDifficulty = [6,5,4,3,2,1], d = 2
输出:7
解释:第一天,您可以完成前 5 项工作,总难度 = 6.
第二天,您可以完成最后一项工作,总难度 = 1.
计划表的难度 = 6 + 1 = 7 

示例 2:

输入:jobDifficulty = [9,9,9], d = 4
输出:-1
解释:就算你每天完成一项工作,仍然有一天是空闲的,你无法制定一份能够满足既定工作时间的计划表。

示例 3:

输入:jobDifficulty = [1,1,1], d = 3
输出:3
解释:工作计划为每天一项工作,总难度为 3 。

示例 4:

输入:jobDifficulty = [7,1,7,1,7,1], d = 3
输出:15

示例 5:

输入:jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
输出:843

 

提示:

  • 1 <= jobDifficulty.length <= 300
  • 0 <= jobDifficulty[i] <= 1000
  • 1 <= d <= 10

解法

方法一:动态规划

我们定义 $f[i][j]$ 表示完成前 $i$ 项工作,且一共用了 $j$ 天的最小难度。初始时 $f[0][0] = 0$,其余 $f[i][j]$ 均为 $\infty$。

考虑第 $j$ 天的工作安排,我们可以枚举第 $j$ 天完成的工作 $[k,..i]$,那么有状态转移方程:

$$ f[i][j] = \min_{k \in [1,i]} {f[k-1][j-1] + \max_{k \leq t \leq i} {jobDifficulty[t-1]}} $$

最终答案即为 $f[n][d]$。

时间复杂度 $O(n^2 \times d)$,空间复杂度 $O(n \times d)$。其中 $n$ 和 $d$ 分别为工作数量和需要计划的天数。

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class Solution:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        n = len(jobDifficulty)
        f = [[inf] * (d + 1) for _ in range(n + 1)]
        f[0][0] = 0
        for i in range(1, n + 1):
            for j in range(1, min(d + 1, i + 1)):
                mx = 0
                for k in range(i, 0, -1):
                    mx = max(mx, jobDifficulty[k - 1])
                    f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx)
        return -1 if f[n][d] >= inf else f[n][d]
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class Solution {
    public int minDifficulty(int[] jobDifficulty, int d) {
        final int inf = 1 << 30;
        int n = jobDifficulty.length;
        int[][] f = new int[n + 1][d + 1];
        for (var g : f) {
            Arrays.fill(g, inf);
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= Math.min(d, i); ++j) {
                int mx = 0;
                for (int k = i; k > 0; --k) {
                    mx = Math.max(mx, jobDifficulty[k - 1]);
                    f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx);
                }
            }
        }
        return f[n][d] >= inf ? -1 : f[n][d];
    }
}
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class Solution {
public:
    int minDifficulty(vector<int>& jobDifficulty, int d) {
        int n = jobDifficulty.size();
        int f[n + 1][d + 1];
        memset(f, 0x3f, sizeof(f));
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= min(d, i); ++j) {
                int mx = 0;
                for (int k = i; k; --k) {
                    mx = max(mx, jobDifficulty[k - 1]);
                    f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx);
                }
            }
        }
        return f[n][d] == 0x3f3f3f3f ? -1 : f[n][d];
    }
};
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func minDifficulty(jobDifficulty []int, d int) int {
    n := len(jobDifficulty)
    f := make([][]int, n+1)
    const inf = 1 << 30
    for i := range f {
        f[i] = make([]int, d+1)
        for j := range f[i] {
            f[i][j] = inf
        }
    }
    f[0][0] = 0
    for i := 1; i <= n; i++ {
        for j := 1; j <= min(d, i); j++ {
            mx := 0
            for k := i; k > 0; k-- {
                mx = max(mx, jobDifficulty[k-1])
                f[i][j] = min(f[i][j], f[k-1][j-1]+mx)
            }
        }
    }
    if f[n][d] == inf {
        return -1
    }
    return f[n][d]
}
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function minDifficulty(jobDifficulty: number[], d: number): number {
    const n = jobDifficulty.length;
    const inf = 1 << 30;
    const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(d + 1).fill(inf));
    f[0][0] = 0;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= Math.min(d, i); ++j) {
            let mx = 0;
            for (let k = i; k > 0; --k) {
                mx = Math.max(mx, jobDifficulty[k - 1]);
                f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx);
            }
        }
    }
    return f[n][d] < inf ? f[n][d] : -1;
}

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