LC.P2050[并行课程III]

方法一：拓扑排序+动态规划

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution {
    public int minimumTime(int n, int[][] relations, int[] time) {
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        int[] in = new int[n];
        for (int[] e : relations) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].add(b);
            ++in[b];
        }
        Deque<Integer> queue = new ArrayDeque<>();
        // f[i] 表示节点 i 的最早完成时间
        int[] f = new int[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int v = in[i], t = time[i];
            if (v == 0) {
                queue.offer(i);
                f[i] = t;
                ans = Math.max(ans, t);
            }
        }
        while (!queue.isEmpty()) {
            int i = queue.poll();
            for (int j : g[i]) {
                f[j] = Math.max(f[j], f[i] + time[j]);
                ans = Math.max(ans, f[j]);
                if (--in[j] == 0) {
                    queue.offer(j);
                }
            }
        }
        return ans;
    }
}
  • 时间复杂度：$O(m + n)$
  • 空间复杂度：$O(m + n)$