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).
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.
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.
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.
implSolution{pubfnminimum_time(n:i32,relations:Vec<Vec<i32>>,time:Vec<i32>) -> i32{letmut next_courses = vec![vec![]; n asusize];letmut indgree = vec![0; n asusize];letmut start = vec![0; n asusize];letmut stack = vec![];letmut ret = 0;for relation in&relations {let(prev, next) = (relation[0]asusize - 1, relation[1]asusize - 1); next_courses[prev].push(next); indgree[next] += 1;}for i in0..n asusize{if indgree[i] == 0{ stack.push(i);}}whileletSome(prev) = stack.pop(){let end = start[prev] + time[prev];for&next in&next_courses[prev]{ indgree[next] -= 1; start[next] = start[next].max(end);if indgree[next] == 0{ stack.push(next);}} ret = ret.max(end);} ret }}