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Strongly Connected Components

Count groups of nodes that can all reach each other.

Cosa imparerai

  • Find groups of mutually reachable nodes
  • Use DFS finishing order or low-link values
  • Work with directed graphs

Problema

Mostra traduzione

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.

Vincoli

1 ≤ n ≤ 100,000; 0 ≤ edges ≤ 200,000

Esempi

Esempio 1
Input:
n = 5
edges = [
    [0, 1],
    [1, 2],
    [2, 0],
    [3, 4]
]

Output atteso:
expected = 3
Input:
long long n = 5;
std::vector<std::vector<long long>> edges = {
    {0, 1},
    {1, 2},
    {2, 0},
    {3, 4}
};

Output atteso:
long long expected = 3;
Input:
long n = 5;
long[][] edges = new long[][] {
    new long[] {0, 1},
    new long[] {1, 2},
    new long[] {2, 0},
    new long[] {3, 4}
};

Output atteso:
long expected = 3;

Spiegazione

0, 1, 2 form a cycle (one component); 3 and 4 can't reach each other, so they are separate: 3.

Esempio 2
Input:
n = 3
edges = [
    [0, 1],
    [1, 0],
    [1, 2],
    [2, 1]
]

Output atteso:
expected = 1
Input:
long long n = 3;
std::vector<std::vector<long long>> edges = {
    {0, 1},
    {1, 0},
    {1, 2},
    {2, 1}
};

Output atteso:
long long expected = 1;
Input:
long n = 3;
long[][] edges = new long[][] {
    new long[] {0, 1},
    new long[] {1, 0},
    new long[] {1, 2},
    new long[] {2, 1}
};

Output atteso:
long expected = 1;

Spiegazione

All three nodes reach each other: 1.

Come affrontarlo

Prova prima da solo, poi apri un suggerimento alla volta.

Suggerimento 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.

Suggerimento 2

Use a graph traversal order on the original graph, then explore the reversed graph to collect one region at a time.

Suggerimento 3

Watch for finishing order and isolated nodes; every node must be assigned exactly once, even if it has no incoming or outgoing edges.

Complessità attesa

O(n + m) time, O(n + m) extra space

Riferimenti

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