c# - C# 中的不相交集实现

Question

我没有在 C# 中使用按等级实现联合 找到不相交集 的任何良好实现，所以我实现了自己的。它以 O(log n) 时间复杂度工作。

是否有更快的(或 C# 中的内置实现)或者我可以使用自己的实现？

class DisjointSetUBR
{
    int[] parent;
    int[] rank; // height of tree

    public DisjointSetUBR(int[] arr)
    {
        parent = new int[arr.Length +1];
        rank = new int[arr.Length + 1];
    }

    public void MakeSet(int i)
    {
        parent[i] = i;
    }

    public int Find(int i)
    {
        while (i!=parent[i]) // If i is not root of tree we set i to his parent until we reach root (parent of all parents)
        {
            i = parent[i]; 
        }
        return i;
    }

    // Path compression, O(log*n). For practical values of n, log* n <= 5
    public int FindPath(int i)
    {
        if (i!=parent[i])
        {
            parent[i] = FindPath(parent[i]);
        }
        return parent[i];
    }

    public void Union(int i, int j)
    {
        int i_id = Find(i); // Find the root of first tree (set) and store it in i_id
        int j_id = Find(j); // // Find the root of second tree (set) and store it in j_id

        if (i_id == j_id) // If roots are equal (they have same parents) than they are in same tree (set)
        {
            return;
        }

        if (rank[i_id] > rank[j_id]) // If height of first tree is larger than second tree
        {
            parent[j_id] = i_id; // We hang second tree under first, parent of second tree is same as first tree
        }
        else
        {
            parent[i_id] = j_id; // We hang first tree under second, parent of first tree is same as second tree
            if (rank[i_id] == rank[j_id]) // If heights are same
            {
                rank[j_id]++; // We hang first tree under second, that means height of tree is incremented by one
            }
        }
    }
}

Answer 1

1. 在 Sedgewick 教授的“算法”一书中，他 mentions “加权快速联合与路径压缩”，它应该具有反阿克曼函数摊销查找/联合的时间。

2. 你是对的，.Net 中没有任何 Disjoint Set 的实现。