Problème
In a directed graph with nodes 0..n−1, a strongly connected component is a maximal group of nodes where every node can reach every other. Count the components (Kosaraju's or Tarjan's algorithm).
Implement count_scc(n, edges).
Input
n: The number of nodes. edges: Directed edges as [from, to].
Output
Return the number of components.
Contraintes
1 ≤ n ≤ 100,000; 0 ≤ edges ≤ 200,000
Exemples
Entrée :
n = 5
edges = [
[0, 1],
[1, 2],
[2, 0],
[3, 4]
]
Sortie attendue :
expected = 3Entrée :
long long n = 5;
std::vector<std::vector<long long>> edges = {
{0, 1},
{1, 2},
{2, 0},
{3, 4}
};
Sortie attendue :
long long expected = 3;Entrée :
long n = 5;
long[][] edges = new long[][] {
new long[] {0, 1},
new long[] {1, 2},
new long[] {2, 0},
new long[] {3, 4}
};
Sortie attendue :
long expected = 3;Explication
0, 1, 2 form a cycle (one component); 3 and 4 can't reach each other, so they are separate: 3.
Entrée :
n = 3
edges = [
[0, 1],
[1, 0],
[1, 2],
[2, 1]
]
Sortie attendue :
expected = 1Entrée :
long long n = 3;
std::vector<std::vector<long long>> edges = {
{0, 1},
{1, 0},
{1, 2},
{2, 1}
};
Sortie attendue :
long long expected = 1;Entrée :
long n = 3;
long[][] edges = new long[][] {
new long[] {0, 1},
new long[] {1, 0},
new long[] {1, 2},
new long[] {2, 1}
};
Sortie attendue :
long expected = 1;Explication
All three nodes reach each other: 1.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
A node belongs with another only if each can reach the other, so think in terms of mutually reachable regions rather than simple reachability.
Indice 2
Use a graph traversal order on the original graph, then explore the reversed graph to collect one region at a time.
Indice 3
Watch for finishing order and isolated nodes; every node must be assigned exactly once, even if it has no incoming or outgoing edges.
Complexité attendue
O(n + m) time, O(n + m) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.