simplex.cpp

// Two-phase simplex algorithm for solving linear programs
// of the form
//
//     maximize     c^T x
//     subject to   Ax <= b
//                  x >= 0
//
// INPUT: A -- an m x n matrix
//        b -- an m-dimensional vector
//        c -- an n-dimensional vector
//        x -- a vector where the optimal solution will be
//        stored
//
// OUTPUT: value of the optimal solution (infinity if
// unbounded
//         above, nan if infeasible)
//
// To use this code, create an LPSolver object with A, b,
// and c as arguments.  Then, call Solve(x).

typedef long double DOUBLE;
typedef vector<DOUBLE> VD;
typedef vector<VD> VVD;
typedef vector<int> VI;

const DOUBLE EPS = 1e-9;

struct LPSolver {
  int m, n;
  VI B, N;
  VVD D;

  LPSolver(const VVD &A, const VD &b, const VD &c)
      : m(b.size()), n(c.size()), N(n + 1), B(m),
        D(m + 2, VD(n + 2)) {
    for (int i = 0; i < m; i++)
      for (int j = 0; j < n; j++)
        D[i][j] = A[i][j];
    for (int i = 0; i < m; i++) {
      B[i] = n + i;
      D[i][n] = -1;
      D[i][n + 1] = b[i];
    }
    for (int j = 0; j < n; j++) {
      N[j] = j;
      D[m][j] = -c[j];
    }
    N[n] = -1;
    D[m + 1][n] = 1;
  }

  void Pivot(int r, int s) {
    DOUBLE inv = 1.0 / D[r][s];
    for (int i = 0; i < m + 2; i++)
      if (i != r)
        for (int j = 0; j < n + 2; j++)
          if (j != s)
            D[i][j] -= D[r][j] * D[i][s] * inv;
    for (int j = 0; j < n + 2; j++)
      if (j != s)
        D[r][j] *= inv;
    for (int i = 0; i < m + 2; i++)
      if (i != r)
        D[i][s] *= -inv;
    D[r][s] = inv;
    swap(B[r], N[s]);
  }

  bool Simplex(int phase) {
    int x = phase == 1 ? m + 1 : m;
    while (true) {
      int s = -1;
      for (int j = 0; j <= n; j++) {
        if (phase == 2 && N[j] == -1)
          continue;
        if (s == -1 || D[x][j] < D[x][s] ||
            D[x][j] == D[x][s] && N[j] < N[s])
          s = j;
      }
      if (s < 0 || D[x][s] > -EPS)
        return true;
      int r = -1;
      for (int i = 0; i < m; i++) {
        if (D[i][s] < EPS)
          continue;
        if (r == -1 ||
            D[i][n + 1] / D[i][s] < D[r][n + 1] / D[r][s] ||
            D[i][n + 1] / D[i][s] == D[r][n + 1] / D[r][s] &&
                B[i] < B[r])
          r = i;
      }
      if (r == -1)
        return false;
      Pivot(r, s);
    }
  }

  DOUBLE Solve(VD &x) {
    int r = 0;
    for (int i = 1; i < m; i++)
      if (D[i][n + 1] < D[r][n + 1])
        r = i;
    if (D[r][n + 1] <= -EPS) {
      Pivot(r, n);
      if (!Simplex(1) || D[m + 1][n + 1] < -EPS)
        return -numeric_limits<DOUBLE>::infinity();
      for (int i = 0; i < m; i++)
        if (B[i] == -1) {
          int s = -1;
          for (int j = 0; j <= n; j++)
            if (s == -1 || D[i][j] < D[i][s] ||
                D[i][j] == D[i][s] && N[j] < N[s])
              s = j;
          Pivot(i, s);
        }
    }
    if (!Simplex(2))
      return numeric_limits<DOUBLE>::infinity();
    x = VD(n);
    for (int i = 0; i < m; i++)
      if (B[i] < n)
        x[B[i]] = D[i][n + 1];
    return D[m][n + 1];
  }
};