Question

2050. Parallel Courses III

中文文档

Description

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.

Solutions

Solution 1

class Solution:
  def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
    g = defaultdict(list)
    indeg = [0] * n
    for a, b in relations:
      g[a - 1].append(b - 1)
      indeg[b - 1] += 1
    q = deque()
    f = [0] * n
    ans = 0
    for i, (v, t) in enumerate(zip(indeg, time)):
      if v == 0:
        q.append(i)
        f[i] = t
        ans = max(ans, t)
    while q:
      i = q.popleft()
      for j in g[i]:
        f[j] = max(f[j], f[i] + time[j])
        ans = max(ans, f[j])
        indeg[j] -= 1
        if indeg[j] == 0:
          q.append(j)
    return ans

class Solution {
  public int minimumTime(int n, int[][] relations, int[] time) {
    List<Integer>[] g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    int[] indeg = new int[n];
    for (int[] e : relations) {
      int a = e[0] - 1, b = e[1] - 1;
      g[a].add(b);
      ++indeg[b];
    }
    Deque<Integer> q = new ArrayDeque<>();
    int[] f = new int[n];
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      int v = indeg[i], t = time[i];
      if (v == 0) {
        q.offer(i);
        f[i] = t;
        ans = Math.max(ans, t);
      }
    }
    while (!q.isEmpty()) {
      int i = q.pollFirst();
      for (int j : g[i]) {
        f[j] = Math.max(f[j], f[i] + time[j]);
        ans = Math.max(ans, f[j]);
        if (--indeg[j] == 0) {
          q.offer(j);
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int minimumTime(int n, vector<vector<int>>& relations, vector<int>& time) {
    vector<vector<int>> g(n);
    vector<int> indeg(n);
    for (auto& e : relations) {
      int a = e[0] - 1, b = e[1] - 1;
      g[a].push_back(b);
      ++indeg[b];
    }
    queue<int> q;
    vector<int> f(n);
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      int v = indeg[i], t = time[i];
      if (v == 0) {
        q.push(i);
        f[i] = t;
        ans = max(ans, t);
      }
    }
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      for (int j : g[i]) {
        if (--indeg[j] == 0) {
          q.push(j);
        }
        f[j] = max(f[j], f[i] + time[j]);
        ans = max(ans, f[j]);
      }
    }
    return ans;
  }
};

func minimumTime(n int, relations [][]int, time []int) int {
  g := make([][]int, n)
  indeg := make([]int, n)
  for _, e := range relations {
    a, b := e[0]-1, e[1]-1
    g[a] = append(g[a], b)
    indeg[b]++
  }
  f := make([]int, n)
  q := []int{}
  ans := 0
  for i, v := range indeg {
    if v == 0 {
      q = append(q, i)
      f[i] = time[i]
      ans = max(ans, time[i])
    }
  }
  for len(q) > 0 {
    i := q[0]
    q = q[1:]
    for _, j := range g[i] {
      indeg[j]--
      if indeg[j] == 0 {
        q = append(q, j)
      }
      f[j] = max(f[j], f[i]+time[j])
      ans = max(ans, f[j])
    }
  }
  return ans
}

function minimumTime(n: number, relations: number[][], time: number[]): number {
  const g: number[][] = Array(n)
    .fill(0)
    .map(() => []);
  const indeg: number[] = Array(n).fill(0);
  for (const [a, b] of relations) {
    g[a - 1].push(b - 1);
    ++indeg[b - 1];
  }
  const q: number[] = [];
  const f: number[] = Array(n).fill(0);
  let ans: number = 0;
  for (let i = 0; i < n; ++i) {
    if (indeg[i] === 0) {
      q.push(i);
      f[i] = time[i];
      ans = Math.max(ans, f[i]);
    }
  }
  while (q.length > 0) {
    const i = q.shift()!;
    for (const j of g[i]) {
      f[j] = Math.max(f[j], f[i] + time[j]);
      ans = Math.max(ans, f[j]);
      if (--indeg[j] === 0) {
        q.push(j);
      }
    }
  }
  return ans;
}