算法训练 与1连通的点的个数

主要思想是并查集，不懂的可以先了解下这个算法再来做题就明白了。
c++实现：

#include<iostream>
#include<vector>
using namespace std;
int f[10000];

//找根节点
int find(int x)
{
    if (f[x] != x)
        f[x] = find(f[x]);
    return f[x];                //不要加else!!!!!!!!! 
}

//连接两个集合
void join(int x, int y)
{
    int a = find(x);
    int b = find(y);
    if (a < b)        //判断两点的根节点大小，合并
        f[b] = a;
    else if (a > b)
        f[a] = b;
}

int main()
{
    int n, m;                        //n：点的个数，m：边
    int temp_1, temp_2,count=0;
    cin >> n >> m;
    //extern vector<int> f(n+1);
    for (int i = 1; i <= n; i++)
    {
        f[i] = i;
    }
    for (int i = 1; i <= m; i++)
    {
        cin >> temp_1 >> temp_2;
        join(temp_1, temp_2);
    }
    //与1同属一个根节点即为两点连通
    int jiedian_1 = find(1);
    for (int i = 1; i <= n; i++)
    {
        if (jiedian_1 == find(i))
            count++;
    }
    cout << count << endl;
    return 0;

以下为java实现代码

import java.util.*;
public class Main {
    static int[] pre;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), m = sc.nextInt();
        pre = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            pre[i] = i;
        }
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt(), b = sc.nextInt();
            join(a, b);
        }
        int traget = find(1), count = 0;
        for (int i = 1; i <= n; i++) {
            int x = find(i);
            if (traget == x) {
                count++;
            }
        }
        System.out.println(count);
    }

    public static void join(int x, int y) {
        int fx = find(x), fy = find(y);
        if (fx > fy) {
            pre[fx] = fy;
        } else {
            pre[fy] = fx;
        }
    }

    public static int find(int x) {
        if (pre[x] != x) {
            pre[x] = find(pre[x]);
        }
        return pre[x];
    }
}