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@@ -1,78 +1,59 @@
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// Two-phase simplex algorithm for solving linear programs of the form
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-//
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// maximize c^T x
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// subject to Ax <= b
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// x >= 0
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-//
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// INPUT: A -- an m x n matrix
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// b -- an m-dimensional vector
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// c -- an n-dimensional vector
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// x -- a vector where the optimal solution will be stored
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-//
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// OUTPUT: value of the optimal solution (infinity if unbounded
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// above, nan if infeasible)
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-//
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-// To use this code, create an LPSolver object with A, b, and c as
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-// arguments. Then, call Solve(x).
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-
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-#include <iostream>
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-#include <iomanip>
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-#include <vector>
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-#include <cmath>
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-#include <limits>
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-
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-using namespace std;
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-
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+// USAGE: create an LPSolver object with A, b, and c as
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+// arguments. Then, call Solve(x).
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typedef long double DOUBLE;
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typedef vector<DOUBLE> VD;
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typedef vector<VD> VVD;
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typedef vector<int> VI;
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-
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const DOUBLE EPS = 1e-9;
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-
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struct LPSolver {
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int m, n;
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VI B, N;
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VVD D;
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-
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- LPSolver(const VVD &A, const VD &b, const VD &c) :
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+ LPSolver(const VVD &A, const VD &b, const VD &c) :
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m(b.size()), n(c.size()), N(n+1), B(m), D(m+2, VD(n+2)) {
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for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) D[i][j] = A[i][j];
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for (int i = 0; i < m; i++) { B[i] = n+i; D[i][n] = -1; D[i][n+1] = b[i]; }
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for (int j = 0; j < n; j++) { N[j] = j; D[m][j] = -c[j]; }
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N[n] = -1; D[m+1][n] = 1;
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}
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-
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void Pivot(int r, int s) {
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for (int i = 0; i < m+2; i++) if (i != r)
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for (int j = 0; j < n+2; j++) if (j != s)
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- D[i][j] -= D[r][j] * D[i][s] / D[r][s];
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+ D[i][j] -= D[r][j] * D[i][s] / D[r][s];
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for (int j = 0; j < n+2; j++) if (j != s) D[r][j] /= D[r][s];
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for (int i = 0; i < m+2; i++) if (i != r) D[i][s] /= -D[r][s];
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D[r][s] = 1.0 / D[r][s];
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swap(B[r], N[s]);
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}
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-
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bool Simplex(int phase) {
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int x = phase == 1 ? m+1 : m;
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while (true) {
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int s = -1;
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for (int j = 0; j <= n; j++) {
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- if (phase == 2 && N[j] == -1) continue;
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- if (s == -1 || D[x][j] < D[x][s] || D[x][j] == D[x][s] && N[j] < N[s]) s = j;
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+ if (phase == 2 && N[j] == -1) continue;
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+ if (s == -1 || D[x][j] < D[x][s] || D[x][j] == D[x][s] && N[j] < N[s]) s = j;
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}
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if (D[x][s] >= -EPS) return true;
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int r = -1;
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for (int i = 0; i < m; i++) {
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- if (D[i][s] <= 0) continue;
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- if (r == -1 || D[i][n+1] / D[i][s] < D[r][n+1] / D[r][s] ||
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- D[i][n+1] / D[i][s] == D[r][n+1] / D[r][s] && B[i] < B[r]) r = i;
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+ if (D[i][s] <= 0) continue;
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+ if (r == -1 || D[i][n+1] / D[i][s] < D[r][n+1] / D[r][s] ||
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+ D[i][n+1] / D[i][s] == D[r][n+1] / D[r][s] && B[i] < B[r]) r = i;
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}
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if (r == -1) return false;
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Pivot(r, s);
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}
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}
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-
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DOUBLE Solve(VD &x) {
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int r = 0;
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for (int i = 1; i < m; i++) if (D[i][n+1] < D[r][n+1]) r = i;
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@@ -80,10 +61,10 @@ struct LPSolver {
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Pivot(r, n);
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if (!Simplex(1) || D[m+1][n+1] < -EPS) return -numeric_limits<DOUBLE>::infinity();
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for (int i = 0; i < m; i++) if (B[i] == -1) {
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- int s = -1;
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- for (int j = 0; j <= n; j++)
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- if (s == -1 || D[i][j] < D[i][s] || D[i][j] == D[i][s] && N[j] < N[s]) s = j;
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- Pivot(i, s);
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+ int s = -1;
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+ for (int j = 0; j <= n; j++)
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+ if (s == -1 || D[i][j] < D[i][s] || D[i][j] == D[i][s] && N[j] < N[s]) s = j;
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+ Pivot(i, s);
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}
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}
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if (!Simplex(2)) return numeric_limits<DOUBLE>::infinity();
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@@ -92,32 +73,3 @@ struct LPSolver {
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return D[m][n+1];
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}
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};
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-
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-int main() {
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-
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- const int m = 4;
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- const int n = 3;
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- DOUBLE _A[m][n] = {
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- { 6, -1, 0 },
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- { -1, -5, 0 },
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- { 1, 5, 1 },
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- { -1, -5, -1 }
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- };
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- DOUBLE _b[m] = { 10, -4, 5, -5 };
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- DOUBLE _c[n] = { 1, -1, 0 };
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-
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- VVD A(m);
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- VD b(_b, _b + m);
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- VD c(_c, _c + n);
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- for (int i = 0; i < m; i++) A[i] = VD(_A[i], _A[i] + n);
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-
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- LPSolver solver(A, b, c);
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- VD x;
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- DOUBLE value = solver.Solve(x);
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-
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- cerr << "VALUE: "<< value << endl;
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- cerr << "SOLUTION:";
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- for (size_t i = 0; i < x.size(); i++) cerr << " " << x[i];
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- cerr << endl;
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- return 0;
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-}
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