Disjoint Set Union 模板

Tue, 07 Apr 2026 10:33:41 +0800

Dsu 内部节点下标从 1 开始.

int Find(int u) 返回节点 u 的根节点
void Union(int u, int v) 合并两个节点所在的集合

之后可以加上带权并查集.

#ifndef DSU_H
#define DSU_H

#include <cassert>
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <vector>

using namespace std;

struct Dsu {
  int n{0};
  vector<int> fa{}, sz{};

  explicit Dsu(int n_) : n(n_), fa{vector<int>(n + 1)}, sz{fa} {
    if (n <= 0) {
      ostringstream output;
      output << "n<=0: " << "n=" << n;
      throw invalid_argument(output.str());
    }

    for (int i = 1; i <= n; i++) {
      fa[i] = i;
      sz[i] = 1;
    }
  }

  int Find(int u) {
    assert(u >= 1 && u <= n);
    return u == fa[u] ? u : (fa[u] = Find(fa[u]));
  }

  void Union(int u, int v) {
    assert(u >= 1 && u <= n && v >= 1 && v <= n);
    int ru = Find(u), rv = Find(v);
    if (ru == rv)
      return;
    if (sz[ru] < sz[rv])
      swap(ru, rv);
    fa[rv] = ru;
    sz[ru] += sz[rv];
  }
};

#endif

CodeForces 1559D1 Mocha and Diana (Easy Version)

Wed, 17 Jan 2024 23:22:38 +0800

Mocha and Diana (Easy Version)

一个连通块中的两个点之间不能添加边，所以想到使用并查集维护集合。设第一个人为 $A$，第二个人是 $B$，那么 $A$ 中的两个点 $x, y$ 能够相连的条件是 $B$ 中的 $x, y$ 不在一个连通块中。如果在 $x, y$ 之间添加边，此时 $A, B$ 中的连通块数量都减少 $1$，当 $B$ 的连通块数量是 $1$ 的时候，就不能再添加边了。对于 $x, y$ 来说，其实换成对应的连通块中的其它符合条件的两个点都可以，它们是等价的，最终效果都一样。最终局面是 $B$ 只有一个连通块，$A$ 的连通块至少有一个。因此检查 $A$ 中的所有点对，能连的就放到答案中即可。

// Date: Wed Jan 17 19:55:47 2024

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
template <class T> using pq = priority_queue<T>;
template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353;
const ll INFL = 0x3f3f3f3f'3f3f3f3f;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nemp(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
#define NL cout << '\n';

template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }

template <typename T, typename V> void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ", ";
  __print(x.second);
  cerr << '}';
}
template <typename T> void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x)
    cerr << (f++ ? ", " : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v))
    cerr << ", ";
  _print(v...);
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#define dbg(x...)                                                              \
  cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = [";    \
  _print(x);                                                                   \
  cerr << "\e[39m" << endl;
#else
#define debug1
#define debug2
#define debug3
#define dbg(x...)
#endif

const int N = 1100;

int f[2][N], sz[2][N], n, m1, m2;

void Init() {
  For1(i, 1, n) {
    f[0][i] = f[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
}

int Find(int i, int x) {
  if (f[i][x] == x)
    return x;
  return f[i][x] = Find(i, f[i][x]);
}

void Union(int i, int x, int y) {
  int rx = Find(i, x), ry = Find(i, y);
  if (rx == ry)
    return;
  f[i][ry] = rx;
}

void solve() {
  while (cin >> n >> m1 >> m2) {
    int u, v;
    Init();

    For(i, 0, m1) {
      cin >> u >> v;
      Union(0, u, v);
    }

    For(i, 0, m2) {
      cin >> u >> v;
      Union(1, u, v);
    }

    vector<PII> res;

    For1(i, 1, n) {
      For1(j, i + 1, n) {
        int ri = Find(0, i), rj = Find(0, j);
        if (ri != rj) {
          int ri1 = Find(1, i), rj1 = Find(1, j);

          if (ri1 != rj1) {
            res.pb({i, j});
            Union(1, i, j);
            Union(0, i, j);
          }
        }
      }
    }

    int len = SZ(res);
    cout << len << '\n';
    if (len) {
      for (auto &[x, y] : res) {
        cout << x << ' ' << y << '\n';
      }
    }
  }
}

int main(void) {
#ifdef _DEBUG
  freopen("1559d1.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  int T = 1;

  while (T--) {
    solve();
  }

  return 0;
}

Disjoint Set Union on Xueyang Liu

Disjoint Set Union 模板

CodeForces 1559D1 Mocha and Diana (Easy Version)