We can interpret such a graph also as a weighted graph, where every edge has the weight $1$.

However if the weights are more constrained, we can often do better.

In this article we demonstrate how we can use BFS to solve the SSSP (single-source shortest path) problem in $O(|E|)$, if the weight of each edge is either $0$ or $1$.

You can see the implication graph in the following image:

It is worth paying attention to the property of the implication graph:

if there is an edge $a \Rightarrow b$, then there also is an edge $\lnot b \Rightarrow \lnot a$.

As we can check easily, neither of the four components contain a vertex $x$ and its negation $\lnot x$, therefore the example has a solution.

We will learn in the next paragraphs how to compute a valid assignment, but just for demonstration purposes the solution $a = \text{false}$, $b = \text{false}$, $c = \text{false}$ is given.

Now we construct the algorithm for finding the solution of the 2-SAT problem on the assumption that the solution exists.

- Given a square matrix $A[1..N, 1..N]$, you need to select $N$ elements in it so that exactly one element is selected in each row and column, and the sum of the values of these elements is the smallest.

- There are $N$ orders and $N$ machines. The cost of manufacturing on each machine is known for each order. Only one order can be performed on each machine. It is required to assign all orders to the machines so that the total cost is minimized.

Let's build a bipartite network: there is a source $S$, a drain $T$, in the first part there are $N$ vertices (corresponding to rows of the matrix, or orders), in the second there are also $N$ vertices (corresponding to the columns of the matrix, or machines). Between each vertex $i$ of the first set and each vertex $j$ of the second set, we draw an edge with bandwidth 1 and cost $A_{ij}$. From the source $S$ we draw edges to all vertices $i$ of the first set with bandwidth 1 and cost 0. We draw an edge with bandwidth 1 and cost 0 from each vertex of the second set $j$ to the drain $T$.

We find in the resulting network the maximum flow of the minimum cost. Obviously, the value of the flow will be $N$. Further, for each vertex $i$ of the first segment there is exactly one vertex $j$ of the second segment, such that the flow $F_{ij}$ = 1. Finally, this is a one-to-one correspondence between the vertices of the first segment and the vertices of the second part, which is the solution to the problem (since the found flow has a minimal cost, then the sum of the costs of the selected edges will be the lowest possible, which is the optimality criterion).

The implementation given here is long, it can probably be significantly reduced.

const int INF = 1000 * 1000 * 1000;