/* -*- mode: C++; indent-tabs-mode: nil; -*-

  *

  * This file is a part of LEMON, a generic C++ optimization library.

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  * Copyright (C) 2003-2009

  * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport

  * (Egervary Research Group on Combinatorial Optimization, EGRES).

  *

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  * This software is provided "AS IS" with no warranty of any kind,

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  */

 namespace lemon {

 /**

 \page min_cost_flow Minimum Cost Flow Problem

 \section mcf_def Definition (GEQ form)

 The \e minimum \e cost \e flow \e problem is to find a feasible flow of

 minimum total cost from a set of supply nodes to a set of demand nodes

 in a network with capacity constraints (lower and upper bounds)

 and arc costs.

 Formally, let \f$G=(V,A)\f$ be a digraph, \f$lower: A\rightarrow\mathbf{R}\f$,

 \f$upper: A\rightarrow\mathbf{R}\cup\{+\infty\}\f$ denote the lower and

 upper bounds for the flow values on the arcs, for which

 \f$lower(uv) \leq upper(uv)\f$ must hold for all \f$uv\in A\f$,

 \f$cost: A\rightarrow\mathbf{R}\f$ denotes the cost per unit flow

 on the arcs and \f$sup: V\rightarrow\mathbf{R}\f$ denotes the

 signed supply values of the nodes.

 If \f$sup(u)>0\f$, then \f$u\f$ is a supply node with \f$sup(u)\f$

 supply, if \f$sup(u)<0\f$, then \f$u\f$ is a demand node with

 \f$-sup(u)\f$ demand.

 A minimum cost flow is an \f$f: A\rightarrow\mathbf{R}\f$ solution

 of the following optimization problem.

 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \geq

     sup(u) \quad \forall u\in V \f]

 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

 The sum of the supply values, i.e. \f$\sum_{u\in V} sup(u)\f$ must be

 zero or negative in order to have a feasible solution (since the sum

 of the expressions on the left-hand side of the inequalities is zero).

 It means that the total demand must be greater or equal to the total

 supply and all the supplies have to be carried out from the supply nodes,

 but there could be demands that are not satisfied.

 If \f$\sum_{u\in V} sup(u)\f$ is zero, then all the supply/demand

 constraints have to be satisfied with equality, i.e. all demands

 have to be satisfied and all supplies have to be used.

 \section mcf_algs Algorithms

 LEMON contains several algorithms for solving this problem, for more

 information see \ref min_cost_flow_algs "Minimum Cost Flow Algorithms".

 A feasible solution for this problem can be found using \ref Circulation.

 \section mcf_dual Dual Solution

 The dual solution of the minimum cost flow problem is represented by

 node potentials \f$\pi: V\rightarrow\mathbf{R}\f$.

 An \f$f: A\rightarrow\mathbf{R}\f$ primal feasible solution is optimal

 if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$ node potentials

 the following \e complementary \e slackness optimality conditions hold.

  - For all \f$uv\in A\f$ arcs:

    - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;

    - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;

    - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.

  - For all \f$u\in V\f$ nodes:

    - \f$\pi(u)<=0\f$;

    - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,

      then \f$\pi(u)=0\f$.

 Here \f$cost^\pi(uv)\f$ denotes the \e reduced \e cost of the arc

 \f$uv\in A\f$ with respect to the potential function \f$\pi\f$, i.e.

 \f[ cost^\pi(uv) = cost(uv) + \pi(u) - \pi(v).\f]

 All algorithms provide dual solution (node potentials), as well,

 if an optimal flow is found.

 \section mcf_eq Equality Form

 The above \ref mcf_def "definition" is actually more general than the

 usual formulation of the minimum cost flow problem, in which strict

 equalities are required in the supply/demand contraints.

 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) =

     sup(u) \quad \forall u\in V \f]

 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

 However if the sum of the supply values is zero, then these two problems

 are equivalent.

 The \ref min_cost_flow_algs "algorithms" in LEMON support the general

 form, so if you need the equality form, you have to ensure this additional

 contraint manually.

 \section mcf_leq Opposite Inequalites (LEQ Form)

 Another possible definition of the minimum cost flow problem is

 when there are <em>"less or equal"</em> (LEQ) supply/demand constraints,

 instead of the <em>"greater or equal"</em> (GEQ) constraints.

 \f[ \min\sum_{uv\in A} f(uv) \cdot cost(uv) \f]

 \f[ \sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \leq

     sup(u) \quad \forall u\in V \f]

 \f[ lower(uv) \leq f(uv) \leq upper(uv) \quad \forall uv\in A \f]

 It means that the total demand must be less or equal to the 

 total supply (i.e. \f$\sum_{u\in V} sup(u)\f$ must be zero or

 positive) and all the demands have to be satisfied, but there

 could be supplies that are not carried out from the supply

 nodes.

 The equality form is also a special case of this form, of course.

 You could easily transform this case to the \ref mcf_def "GEQ form"

 of the problem by reversing the direction of the arcs and taking the

 negative of the supply values (e.g. using \ref ReverseDigraph and

 \ref NegMap adaptors).

 However \ref NetworkSimplex algorithm also supports this form directly

 for the sake of convenience.

 Note that the optimality conditions for this supply constraint type are

 slightly differ from the conditions that are discussed for the GEQ form,

 namely the potentials have to be non-negative instead of non-positive.

 An \f$f: A\rightarrow\mathbf{R}\f$ feasible solution of this problem

 is optimal if and only if for some \f$\pi: V\rightarrow\mathbf{R}\f$

 node potentials the following conditions hold.

  - For all \f$uv\in A\f$ arcs:

    - if \f$cost^\pi(uv)>0\f$, then \f$f(uv)=lower(uv)\f$;

    - if \f$lower(uv)<f(uv)<upper(uv)\f$, then \f$cost^\pi(uv)=0\f$;

    - if \f$cost^\pi(uv)<0\f$, then \f$f(uv)=upper(uv)\f$.

  - For all \f$u\in V\f$ nodes:

    - \f$\pi(u)>=0\f$;

    - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,

      then \f$\pi(u)=0\f$.

 */

 }