union-find algorithm

public class UF
{
	private int[] id; //component id (with the contact as an index)
	private int count; //Number of components
	public UF(int N)
	{
		//initialize the component array
		count = N;
		id = new int[N];
		for (int i = 0; i < N; i++)
			id[i] = i;
	}
	public int count()
	{	return count;	}
	public boolean connected(int p,int q)
	{	return find(p) == find(q);	}
	public int find(int p)
	public void union(int p,int q)
	public static void main(String[] args)
	{
		int N = StdIn.readInt(); //Read the number of contacts
		UF uf = new UF(N); //Initialize N components
		while (!StdIn.isEmpty())
		{
			int p = StdIn.readInt();
			int q = StdIn.readInt(); //read integer pair
			if (uf.connected(p, q)) continue; //Ignore if already connected
			uf.union(p, q); // merge components
			StdOut.println(p + " " + q); //print connection
		}
		StdOut.println(uf.count() + "components");
	}
}

　　

public int find(int p)
{	return id[p];	}
public void union(int p, int q)
{ // merge p and q into the same component
	int pID = find(p);
	int qID = find(q);
	//p and q are in the same component, no action is taken
	if (pID == qID) return;
	//Rename the component of p to the name of q
	for (int i = 0; i < id.length; i++)
		if (id[i] == pID) id[i] = qID;
	count--;
}