graph/maximum_independent_set.hpp

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• Last update: 2024-04-19 02:20:22+09:00
• Include: #include "graph/maximum_independent_set.hpp"
• Link: https://codeforces.com/contest/1578/problem/K

Depends on

• graph/base.hpp

Verified with

• test/library_checker/graph/maximum_independent_set.test.cpp

Code

#include "graph/base.hpp"


// worst N*1.381^N だが、N=100 程度でも使える可能性がある。

// 特殊最大80頂点爆速 https://codeforces.com/contest/1578/problem/K

template <int MAX_V, typename GT>
vc<int> maximum_independent_set(GT& G) {
  using BS = bitset<MAX_V>;
  const int N = G.N;
  assert(N <= MAX_V);
  vc<BS> nbd(N);
  FOR(v, N) for (auto&& e: G[v]) nbd[v][e.to] = 1;

  int best = 0;
  BS res;

  auto dfs = [&](auto& dfs, BS now, BS rest) -> void {
    pair<int, int> p = {-1, -1}; // (v, d)

    while (1) {
      bool upd = 0;
      FOR(v, N) if (rest[v]) {
        int d = (nbd[v] & rest).count();
        if (chmax(p.se, d)) p.fi = v;
        if (d <= 1) { rest[v] = 0, rest &= ~nbd[v], now[v] = 1, upd = 1; }
      }
      if (!upd) break;
      p = {-1, -1};
    }
    int a = now.count(), b = rest.count();
    if (chmax(best, a)) res = now;
    if (b == 0 || a + b <= best) return;
    int v = p.fi;
    rest[v] = 0;
    if (p.se >= 3) dfs(dfs, now, rest);
    now[v] = 1;
    dfs(dfs, now, rest & ~(nbd[v]));
  };
  BS now, rest;
  FOR(v, N) rest[v] = 1;
  dfs(dfs, now, rest);
  vc<int> ANS;
  FOR(v, N) if (res[v]) ANS.eb(v);
  return ANS;
}

#line 2 "graph/base.hpp"

template <typename T>
struct Edge {
  int frm, to;
  T cost;
  int id;
};

template <typename T = int, bool directed = false>
struct Graph {
  static constexpr bool is_directed = directed;
  int N, M;
  using cost_type = T;
  using edge_type = Edge<T>;
  vector<edge_type> edges;
  vector<int> indptr;
  vector<edge_type> csr_edges;
  vc<int> vc_deg, vc_indeg, vc_outdeg;
  bool prepared;

  class OutgoingEdges {
  public:
    OutgoingEdges(const Graph* G, int l, int r) : G(G), l(l), r(r) {}

    const edge_type* begin() const {
      if (l == r) { return 0; }
      return &G->csr_edges[l];
    }

    const edge_type* end() const {
      if (l == r) { return 0; }
      return &G->csr_edges[r];
    }

  private:
    const Graph* G;
    int l, r;
  };

  bool is_prepared() { return prepared; }

  Graph() : N(0), M(0), prepared(0) {}
  Graph(int N) : N(N), M(0), prepared(0) {}

  void build(int n) {
    N = n, M = 0;
    prepared = 0;
    edges.clear();
    indptr.clear();
    csr_edges.clear();
    vc_deg.clear();
    vc_indeg.clear();
    vc_outdeg.clear();
  }

  void add(int frm, int to, T cost = 1, int i = -1) {
    assert(!prepared);
    assert(0 <= frm && 0 <= to && to < N);
    if (i == -1) i = M;
    auto e = edge_type({frm, to, cost, i});
    edges.eb(e);
    ++M;
  }

#ifdef FASTIO
  // wt, off
  void read_tree(bool wt = false, int off = 1) { read_graph(N - 1, wt, off); }

  void read_graph(int M, bool wt = false, int off = 1) {
    for (int m = 0; m < M; ++m) {
      INT(a, b);
      a -= off, b -= off;
      if (!wt) {
        add(a, b);
      } else {
        T c;
        read(c);
        add(a, b, c);
      }
    }
    build();
  }
#endif

  void build() {
    assert(!prepared);
    prepared = true;
    indptr.assign(N + 1, 0);
    for (auto&& e: edges) {
      indptr[e.frm + 1]++;
      if (!directed) indptr[e.to + 1]++;
    }
    for (int v = 0; v < N; ++v) { indptr[v + 1] += indptr[v]; }
    auto counter = indptr;
    csr_edges.resize(indptr.back() + 1);
    for (auto&& e: edges) {
      csr_edges[counter[e.frm]++] = e;
      if (!directed)
        csr_edges[counter[e.to]++] = edge_type({e.to, e.frm, e.cost, e.id});
    }
  }

  OutgoingEdges operator[](int v) const {
    assert(prepared);
    return {this, indptr[v], indptr[v + 1]};
  }

  vc<int> deg_array() {
    if (vc_deg.empty()) calc_deg();
    return vc_deg;
  }

  pair<vc<int>, vc<int>> deg_array_inout() {
    if (vc_indeg.empty()) calc_deg_inout();
    return {vc_indeg, vc_outdeg};
  }

  int deg(int v) {
    if (vc_deg.empty()) calc_deg();
    return vc_deg[v];
  }

  int in_deg(int v) {
    if (vc_indeg.empty()) calc_deg_inout();
    return vc_indeg[v];
  }

  int out_deg(int v) {
    if (vc_outdeg.empty()) calc_deg_inout();
    return vc_outdeg[v];
  }

#ifdef FASTIO
  void debug() {
    print("Graph");
    if (!prepared) {
      print("frm to cost id");
      for (auto&& e: edges) print(e.frm, e.to, e.cost, e.id);
    } else {
      print("indptr", indptr);
      print("frm to cost id");
      FOR(v, N) for (auto&& e: (*this)[v]) print(e.frm, e.to, e.cost, e.id);
    }
  }
#endif

  vc<int> new_idx;
  vc<bool> used_e;

  // G における頂点 V[i] が、新しいグラフで i になるようにする
  // {G, es}
  Graph<T, directed> rearrange(vc<int> V, bool keep_eid = 0) {
    if (len(new_idx) != N) new_idx.assign(N, -1);
    int n = len(V);
    FOR(i, n) new_idx[V[i]] = i;
    Graph<T, directed> G(n);
    vc<int> history;
    FOR(i, n) {
      for (auto&& e: (*this)[V[i]]) {
        if (len(used_e) <= e.id) used_e.resize(e.id + 1);
        if (used_e[e.id]) continue;
        int a = e.frm, b = e.to;
        if (new_idx[a] != -1 && new_idx[b] != -1) {
          history.eb(e.id);
          used_e[e.id] = 1;
          int eid = (keep_eid ? e.id : -1);
          G.add(new_idx[a], new_idx[b], e.cost, eid);
        }
      }
    }
    FOR(i, n) new_idx[V[i]] = -1;
    for (auto&& eid: history) used_e[eid] = 0;
    G.build();
    return G;
  }

private:
  void calc_deg() {
    assert(vc_deg.empty());
    vc_deg.resize(N);
    for (auto&& e: edges) vc_deg[e.frm]++, vc_deg[e.to]++;
  }

  void calc_deg_inout() {
    assert(vc_indeg.empty());
    vc_indeg.resize(N);
    vc_outdeg.resize(N);
    for (auto&& e: edges) { vc_indeg[e.to]++, vc_outdeg[e.frm]++; }
  }
};
#line 2 "graph/maximum_independent_set.hpp"

// worst N*1.381^N だが、N=100 程度でも使える可能性がある。

// 特殊最大80頂点爆速 https://codeforces.com/contest/1578/problem/K

template <int MAX_V, typename GT>
vc<int> maximum_independent_set(GT& G) {
  using BS = bitset<MAX_V>;
  const int N = G.N;
  assert(N <= MAX_V);
  vc<BS> nbd(N);
  FOR(v, N) for (auto&& e: G[v]) nbd[v][e.to] = 1;

  int best = 0;
  BS res;

  auto dfs = [&](auto& dfs, BS now, BS rest) -> void {
    pair<int, int> p = {-1, -1}; // (v, d)

    while (1) {
      bool upd = 0;
      FOR(v, N) if (rest[v]) {
        int d = (nbd[v] & rest).count();
        if (chmax(p.se, d)) p.fi = v;
        if (d <= 1) { rest[v] = 0, rest &= ~nbd[v], now[v] = 1, upd = 1; }
      }
      if (!upd) break;
      p = {-1, -1};
    }
    int a = now.count(), b = rest.count();
    if (chmax(best, a)) res = now;
    if (b == 0 || a + b <= best) return;
    int v = p.fi;
    rest[v] = 0;
    if (p.se >= 3) dfs(dfs, now, rest);
    now[v] = 1;
    dfs(dfs, now, rest & ~(nbd[v]));
  };
  BS now, rest;
  FOR(v, N) rest[v] = 1;
  dfs(dfs, now, rest);
  vc<int> ANS;
  FOR(v, N) if (res[v]) ANS.eb(v);
  return ANS;
}

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