并查集求连通分量个数、优美的并查集板子

 

 

思路：

找祖先的数量，祖先的数量即为连通块的数量，并查集基本操作

 AC代码：
 

import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;

public class Main
{
    static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    static int N = (int)2e5 + 10;
    static mqm bcj = new mqm();
    static Set<Integer> set = new HashSet<>();

    public static void main(String[] args ) throws IOException
    {
        int n = rd.nextInt();
        int m = rd.nextInt();

        bcj.init();
        for(int i = 1 ; i <= m ; i ++)
        {
            int u = rd.nextInt(),v =rd.nextInt();
            bcj.merge(u,v);
        }

        // 找祖先的数量，祖先的数量即为连通块的数量
        // 寻找每个点的祖先，丢到set里面去重，完美
        for(int i = 1 ; i <= n ; i ++)
        {
            set.add(bcj.find(i));
        }
        pw.println(set.size());
        pw.flush();
    }
}

class rd
{
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static StringTokenizer tokenizer = new StringTokenizer("");

    static String nextLine() throws IOException { return reader.readLine(); }
    static String next() throws IOException
    {
        while(!tokenizer.hasMoreTokens())  tokenizer = new StringTokenizer(reader.readLine());
        return tokenizer.nextToken();
    }
    static int nextInt() throws IOException { return Integer.parseInt(next()); }
    static double nextDouble() throws IOException { return Double.parseDouble(next()); }
    static long nextLong() throws IOException { return Long.parseLong(next()); }
    static BigInteger nextBigInteger() throws IOException
    {
        BigInteger d = new BigInteger(rd.nextLine());
        return d;
    }
}

class math
{
    int gcd(int a,int b)
    {
        if(b == 0)  return a;
        else return gcd(b,a % b);
    }

    int lcm(int a,int b)
    {
        return a * b / gcd(a, b);
    }

    // 求n的所有约数
    List get_factor(int n)
    {
        List<Long> a = new ArrayList<>();
        for(long i = 1; i <= Math.sqrt(n) ; i ++)
        {
            if(n % i == 0)
            {
                a.add(i);
                if(i != n / i)  a.add(n / i);  // // 避免一下的情况：x = 16时，i = 4 ，x / i = 4的情况，这样会加入两种情况  ^-^复杂度能减少多少是多少
            }
        }

        // 相同因子去重，这个方法，完美
        a = a.stream().distinct().collect(Collectors.toList());

        // 对因子排序（升序）
        Collections.sort(a);

        return a;
    }

    // 判断是否是质数
    boolean check_isPrime(int n)
    {
        if(n < 2) return false;
        for(int i = 2 ; i <= n / i; i ++)  if (n % i == 0) return false;
        return true;
    }
}

class PII implements Comparable<PII>
{
    int x,y;
    public PII(int x ,int y)
    {
        this.x = x;
        this.y = y;
    }
    public int compareTo(PII a)
    {
        if(this.x-a.x != 0)
            return this.x-a.x;  //按x升序排序
        else return this.y-a.y;  //如果x相同，按y升序排序
    }
}

class Edge
{
    int a,b,c;
    public Edge(int a ,int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }
}

class Line implements Comparable<Line>
{
    double k; // 斜率
    double b; // 截距

    public Line(double k, double b)
    {
        this.k = k;
        this.b = b;
    }

    @Override
    public int compareTo(Line o)
    {
        if (this.k > o.k) return 1;
        if (this.k == o.k)
        {
            if (this.b > o.b) return 1;
            return -1;
        }
        return -1;
    }
}

class mqm
{
    int fa[] = new int[1005];
    void init()
    {
        for(int i = 1 ; i <= 1000 ; i ++)  fa[i] = i;
    }
    void merge(int x, int y) { fa[find(x)] = find(y); }
    int find(int x)
    {
        if(x != fa[x])  fa[x] = find(fa[x]);
        return fa[x];
    }
    boolean query(int x, int y) { return find(x) == find(y); }
}