Hungarian(二部グラフの最小重み最大マッチング) (graph/flow/hungarian.hpp)

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• Last update: 2024-04-27 01:27:28+09:00
• Include: #include "graph/flow/hungarian.hpp"

概要

二部グラフの最小重み最大マッチングをハンガリアン法により求める. 特に $N=M$ (左側の頂点数と右側の頂点数が一致する) の場合, 割当問題とも言われる.

使い方

• hungarian(A): 重み行列 A の最小重み最大マッチングを返す. first にそのときの重みの和, second が各行がマッチングした番号を格納して返す. 最大重みを求めたいときは重み行列の重みをすべて $-1$ 倍する. 行列 $A$ は 1-indexed で $N \leq M$ を満たす必要がある(つまり $0$ 行目または $0$ 列目は全て $0$ 埋めする).

計算量

• $O(N^2 M)$

Depends on

• math/matrix/matrix.hpp

Verified with

• test/verify/aoj-1163.test.cpp
• test/verify/yosupo-assignment.test.cpp

Code

#include "../../math/matrix/matrix.hpp"

/**
 * @brief Hungarian(二部グラフの最小重み最大マッチング)
 *
 */
template <typename T>
pair<T, vector<int> > hungarian(Matrix<T> &A) {
  const T infty = numeric_limits<T>::max();
  const int N = (int)A.height();
  const int M = (int)A.width();
  vector<int> P(M), way(M);
  vector<T> U(N, 0), V(M, 0), minV;
  vector<bool> used;

  for (int i = 1; i < N; i++) {
    P[0] = i;
    minV.assign(M, infty);
    used.assign(M, false);
    int j0 = 0;
    while (P[j0] != 0) {
      int i0 = P[j0], j1 = 0;
      used[j0] = true;
      T delta = infty;
      for (int j = 1; j < M; j++) {
        if (used[j]) continue;
        T curr = A[i0][j] - U[i0] - V[j];
        if (curr < minV[j]) minV[j] = curr, way[j] = j0;
        if (minV[j] < delta) delta = minV[j], j1 = j;
      }
      for (int j = 0; j < M; j++) {
        if (used[j])
          U[P[j]] += delta, V[j] -= delta;
        else
          minV[j] -= delta;
      }
      j0 = j1;
    }
    do {
      P[j0] = P[way[j0]];
      j0 = way[j0];
    } while (j0 != 0);
  }
  return {-V[0], P};
}

#line 1 "math/matrix/matrix.hpp"
template <class T>
struct Matrix {
  vector<vector<T> > A;

  Matrix() {}

  Matrix(size_t n, size_t m) : A(n, vector<T>(m, 0)) {}

  Matrix(size_t n) : A(n, vector<T>(n, 0)) {};

  size_t size() const {
    if (A.empty()) return 0;
    assert(A.size() == A[0].size());
    return A.size();
  }

  size_t height() const { return (A.size()); }

  size_t width() const { return (A[0].size()); }

  inline const vector<T> &operator[](int k) const { return (A.at(k)); }

  inline vector<T> &operator[](int k) { return (A.at(k)); }

  static Matrix I(size_t n) {
    Matrix mat(n);
    for (int i = 0; i < n; i++) mat[i][i] = 1;
    return (mat);
  }

  Matrix &operator+=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) (*this)[i][j] += B[i][j];
    return (*this);
  }

  Matrix &operator-=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) (*this)[i][j] -= B[i][j];
    return (*this);
  }

  Matrix &operator*=(const Matrix &B) {
    size_t n = height(), m = B.width(), p = width();
    assert(p == B.height());
    vector<vector<T> > C(n, vector<T>(m, 0));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        for (int k = 0; k < p; k++)
          C[i][j] = (C[i][j] + (*this)[i][k] * B[k][j]);
    A.swap(C);
    return (*this);
  }

  Matrix &operator^=(long long k) {
    Matrix B = Matrix::I(height());
    while (k > 0) {
      if (k & 1) B *= *this;
      *this *= *this;
      k >>= 1LL;
    }
    A.swap(B.A);
    return (*this);
  }

  Matrix operator+(const Matrix &B) const { return (Matrix(*this) += B); }

  Matrix operator-(const Matrix &B) const { return (Matrix(*this) -= B); }

  Matrix operator*(const Matrix &B) const { return (Matrix(*this) *= B); }

  Matrix operator^(const long long k) const { return (Matrix(*this) ^= k); }

  friend ostream &operator<<(ostream &os, Matrix &p) {
    size_t n = p.height(), m = p.width();
    for (int i = 0; i < n; i++) {
      os << "[";
      for (int j = 0; j < m; j++) {
        os << p[i][j] << (j + 1 == m ? "]\n" : ",");
      }
    }
    return (os);
  }

  T determinant() {
    Matrix B(*this);
    assert(width() == height());
    T ret = 1;
    for (int i = 0; i < width(); i++) {
      int idx = -1;
      for (int j = i; j < width(); j++) {
        if (B[j][i] != 0) idx = j;
      }
      if (idx == -1) return (0);
      if (i != idx) {
        ret *= -1;
        swap(B[i], B[idx]);
      }
      ret *= B[i][i];
      T vv = B[i][i];
      for (int j = 0; j < width(); j++) {
        B[i][j] /= vv;
      }
      for (int j = i + 1; j < width(); j++) {
        T a = B[j][i];
        for (int k = 0; k < width(); k++) {
          B[j][k] -= B[i][k] * a;
        }
      }
    }
    return (ret);
  }
};
#line 2 "graph/flow/hungarian.hpp"

/**
 * @brief Hungarian(二部グラフの最小重み最大マッチング)
 *
 */
template <typename T>
pair<T, vector<int> > hungarian(Matrix<T> &A) {
  const T infty = numeric_limits<T>::max();
  const int N = (int)A.height();
  const int M = (int)A.width();
  vector<int> P(M), way(M);
  vector<T> U(N, 0), V(M, 0), minV;
  vector<bool> used;

  for (int i = 1; i < N; i++) {
    P[0] = i;
    minV.assign(M, infty);
    used.assign(M, false);
    int j0 = 0;
    while (P[j0] != 0) {
      int i0 = P[j0], j1 = 0;
      used[j0] = true;
      T delta = infty;
      for (int j = 1; j < M; j++) {
        if (used[j]) continue;
        T curr = A[i0][j] - U[i0] - V[j];
        if (curr < minV[j]) minV[j] = curr, way[j] = j0;
        if (minV[j] < delta) delta = minV[j], j1 = j;
      }
      for (int j = 0; j < M; j++) {
        if (used[j])
          U[P[j]] += delta, V[j] -= delta;
        else
          minV[j] -= delta;
      }
      j0 = j1;
    }
    do {
      P[j0] = P[way[j0]];
      j0 = way[j0];
    } while (j0 != 0);
  }
  return {-V[0], P};
}

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