2050. Parallel Courses III

Created: 2021/10/25

Leetcode

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course.

You must find the minimum number of months needed to complete all the courses following these rules:

• You may start taking a course at any time if the prerequisites are met.

• Any number of courses can be taken at the same time.

Return the minimum number of months needed to complete all the courses.

Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

Example 1:

Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course. 
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.

Example 2:

Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.

Constraints:

• 1 <= n <= 5 * 104

• 0 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)

• relations[j].length == 2

• 1 <= prevCoursej, nextCoursej <= n

• prevCoursej != nextCoursej

• All the pairs [prevCoursej, nextCoursej] are unique.

• time.length == n

• 1 <= time[i] <= 104

• The given graph is a directed acyclic graph.

Basic Idea

每门课都需要在所有先修课上完之后才能上，而同时可以有任意多的课一起上。整体的逻辑还是一个topological sort，但我们仍然需要想办法追踪每门课最终完成需要的时间，最终返回的就是所有课程完成时间中最长的一个。那么如何追踪每门课完成的时间呢？只要用另一个map来记录每门课的先修课中结束最晚的时间加上该课自己的时间即可。

Java Code：

class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        int[] indegree = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            graph.put(i, new ArrayList<>());
        }
        for (int[] edge : relations) {
            graph.get(edge[0]).add(edge[1]);
            indegree[edge[1]]++;
        }
        Queue<Integer> queue = new ArrayDeque<>();
        int[] maxTime = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            if (indegree[i] == 0) {
                queue.offer(i);
                // 没有先修课的课程时间初始化为自己的时间
                maxTime[i] = time[i - 1]; 
            }
        }
        
        // 最晚完成的课程所需时间
        int ret = 0;
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            ret = Math.max(ret, time[curr - 1]);
            for (int next : graph.get(curr)) {
                indegree[next]--;
                // 每个课程的完成时间为最晚完成的先修课时间加上自己的时间
                maxTime[next] = Math.max(maxTime[next], maxTime[curr] + time[next - 1]);
                ret = Math.max(ret, maxTime[next]);
                if (indegree[next] == 0) {
                    queue.offer(next);
                }
            }
        }
        return ret;
    }
}