并查集模板+应用（图文并茂，保你看懂）

#include <iostream>
using namespace std;
const int N = 100010;
int p[N];
int find(int x)//返回祖宗结点
{
    if (p[x] != x) p[x] = find(p[x]);//硬使p[x]=它的祖宗结点
    return p[x];
}
int main()
{
    int n, m;
    cin>>n>>m;
    for (int i = 1; i <= n; i ++ ) p[i] = i;//i是p[i]的父结点
    while (m -- )
    {
        char op;
        int a, b;
        cin>>op>>a>>b;
        if (op == 'M') p[find(a)] = find(b);//find[a]:a的祖宗结点find[b]:b的祖宗结点
        else //也可以是op[0]                  //让a的祖宗结点的父节点指向b的祖宗结点
        {                                    //就是把a，b合并
            if (find(a) == find(b))
            puts("Yes");//如果a，b的祖宗结点一样，证明a，b在一个集合中
            else puts("No");
        }
    }
    return 0;
}

#include <iostream>
using namespace std;
const int N = 100010;
int n, m;
int p[N], cnt[N];
int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ )
    {
        p[i] = i;
        cnt[i] = 1;
    }
    while (m -- )
    {
        string op;
        int a, b;
        cin >> op;
        if (op == "C")
        {
            cin >> a >> b;
            a = find(a), b = find(b);
            if (a != b)
            {
                p[a] = b;
                cnt[b] += cnt[a];
            }
        }
        else if (op == "Q1")
        {
            cin >> a >> b;
            if (find(a) == find(b)) puts("Yes");
            else puts("No");
        }
        else
        {
            cin >> a;
            cout << cnt[find(a)] << endl;
        }
    }
    return 0;
}

#include <iostream>
using namespace std;
const int N= 1e5 + 10;
struct node
{
    int xi, yi;
} w[N];
int p[N];
int find(int x) //路径压缩
{
    if (p[x]!=x)  p[x]=find(p[x]);
    return p[x];
}
void merge(int x, int y)
{
    int a = find(x);
    int b = find(y);
    if (a != b)
        p[a] = b;
}
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) p[i] = i;//初始化
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &w[i].xi, &w[i].yi);
    for (int i = 1; i < n; i++) //防止越界，i要小于n不能等
    {
      for (int j = i + 1; j <= n; j++)
        {
          if (w[i].xi == w[j].xi || w[i].yi == w[j].yi)//横坐标相等||纵坐标相等
                merge(i, j);//合并为一个联通块
                //注意：这里合并的是i,j 不是w[i]w[j]
                //因为w[i]w[j]仅仅是判断横坐标纵坐标是否相等
                //⭐⭐⭐i,j来组成联通块⭐⭐⭐
          //if(find(i)!=find(j))
          //p[find(i)]=find(j);
    }
  }
  //分散的点连成联通块后，判断有多少个联通块
    int ans = 0;
    for (int i = 1; i <= n; ++i)
        if (p[i] == i)
            ans++;
    printf("%d\n", ans - 1);
}