diff -r 7c4ba7daaf5f -r 2b6bffe0e7e8 doc/min_cost_flow.dox
--- a/doc/min_cost_flow.dox	Tue Dec 20 17:44:38 2011 +0100
+++ b/doc/min_cost_flow.dox	Tue Dec 20 18:15:14 2011 +0100
@@ -2,7 +2,7 @@
  *
  * This file is a part of LEMON, a generic C++ optimization library.
  *
- * Copyright (C) 2003-2009
+ * Copyright (C) 2003-2010
  * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport
  * (Egervary Research Group on Combinatorial Optimization, EGRES).
  *
@@ -26,7 +26,7 @@
 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.
+and arc costs \ref amo93networkflows.
 
 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
@@ -78,10 +78,10 @@
    - 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$\pi(u)<=0\f$;
+   - \f$\pi(u)\leq 0\f$;
    - 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]
@@ -119,7 +119,7 @@
     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 
+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
@@ -145,7 +145,7 @@
    - 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$\pi(u)>=0\f$;
+   - \f$\pi(u)\geq 0\f$;
    - if \f$\sum_{uv\in A} f(uv) - \sum_{vu\in A} f(vu) \neq sup(u)\f$,
      then \f$\pi(u)=0\f$.