## Algoritmo de Simplex.

Cualquier duda no dudes en contactar.

``````/**
* Author: Stanford
* Source: Stanford Notebook
* Description: Solves a general linear maximization problem: maximize \$c^T x\$ subject to \$Ax \le b\$, \$x \ge 0\$.
* Returns -inf if there is no solution, inf if there are arbitrarily good solutions, or the maximum value of \$c^T x\$ otherwise.
* The input vector is set to an optimal \$x\$ (or in the unbounded case, an arbitrary solution fulfilling the constraints).
* Numerical stability is not guaranteed. For better performance, define variables such that \$x = 0\$ is viable.
* Usage:
* vvd A = {{1,-1}, {-1,1}, {-1,-2}};
* vd b = {1,1,-4}, c = {-1,-1}, x;
* T val = LPSolver(A, b, c).solve(x);
* Time: O(NM * \#pivots), where a pivot may be e.g. an edge relaxation. O(2^n) in the general case.
* Status: seems to work?
*/
#pragma once

typedef double T; // long double, Rational, double + mod<P>...
typedef vector<T> vd;
typedef vector<vd> vvd;

const T eps = 1e-8, inf = 1/.0;
#define MP make_pair
#define ltj(X) if(s == -1 || MP(X[j],N[j]) < MP(X[s],N[s])) s=j

struct LPSolver {
int m, n;
vi N, B;
vvd D;

LPSolver(const vvd& A, const vd& b, const vd& c) :
m(sz(b)), n(sz(c)), N(n+1), B(m), D(m+2, vd(n+2)) {
FOR(i,0,m) FOR(j,0,n) D[i][j] = A[i][j];
FOR(i,0,m) { B[i] = n+i; D[i][n] = -1; D[i][n+1] = b[i];}
FOR(j,0,n) { N[j] = j; D[m][j] = -c[j]; }
N[n] = -1; D[m+1][n] = 1;
}

void pivot(int r, int s) {
T *a = D[r].data(), inv = 1 / a[s];
FOR(i,0,m+2) if (i != r && abs(D[i][s]) > eps) {
T *b = D[i].data(), inv2 = b[s] * inv;
FOR(j,0,n+2) b[j] -= a[j] * inv2;
b[s] = a[s] * inv2;
}
FOR(j,0,n+2) if (j != s) D[r][j] *= inv;
FOR(i,0,m+2) if (i != r) D[i][s] *= -inv;
D[r][s] = inv;
swap(B[r], N[s]);
}

bool simplex(int phase) {
int x = m + phase - 1;
for (;;) {
int s = -1;
FOR(j,0,n+1) if (N[j] != -phase) ltj(D[x]);
if (D[x][s] >= -eps) return true;
int r = -1;
FOR(i,0,m) {
if (D[i][s] <= eps) continue;
if (r == -1 || MP(D[i][n+1] / D[i][s], B[i])
< MP(D[r][n+1] / D[r][s], B[r])) r = i;
}
if (r == -1) return false;
pivot(r, s);
}
}

T solve(vd &x) {
int r = 0;
FOR(i,1,m) if (D[i][n+1] < D[r][n+1]) r = i;
if (D[r][n+1] < -eps) {
pivot(r, n);
if (!simplex(2) || D[m+1][n+1] < -eps) return -inf;
FOR(i,0,m) if (B[i] == -1) {
int s = 0;
FOR(j,1,n+1) ltj(D[i]);
pivot(i, s);
}
}
bool ok = simplex(1); x = vd(n);
FOR(i,0,m) if (B[i] < n) x[B[i]] = D[i][n+1];
return ok ? D[m][n+1] : inf;
}
};
``````