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Let the bipartite graph have parts $L,R$. Add a source $s$, a sink $t$, capacity-one edges $s\to u$ for $u\in L$ and $v\to t$ for $v\in R$, and capacity $M=|V|+1$ on each original edge directed $L\to R$. Use the max-flow min-cut theorem: maximum flow value equals minimum cut capacity. Also use Integrality of the Ford-Fulkerson algorithm: integral capacities admit an integral maximum flow, found by integral augmenting paths.

An integral flow selects a matching in a graph, because each left and right vertex carries at most one unit. Conversely every matching gives a unit flow along its selected $s\to u\to v\to t$ paths. Hence maximum flow value equals maximum matching cardinality.

A minimum cut has capacity at most $|L|$, whereas crossing even one original edge would cost $M>|L|$. For its source side $Z$, no edge of the original graph runs from $L\cap Z$ to $R\setminus Z$. Thereforeis a vertex cover, and its cardinality is exactly the cut capacity. Conversely, given any cover $U$, use $Z=\{s\}\cup (L\setminus U)\cup (R\cap U)$. No original edge crosses this cut, because both its endpoints would otherwise be outside the cover; its capacity is $|U|$. Thus minimum cut capacity equals minimum cover size, proving König's theorem for bipartite matching:For the algorithm, repeatedly find augmenting paths in the residual network, then take $Z$ to be the vertices reachable from $s$. Each integral augmentation increases flow by at least one; the total value is at most $\min (|L|,|R|)$. Each reachability search takes $O(|V|+|E|)$ time, so this procedure takes $O(|V|(|V|+|E|))$ time. The network uses only polynomially many edges and integer capacities, yielding the requested polynomial-time algorithm and an explicit cover.

A cut is a partition $(S,T)$ of the vertices with the source in $S$ and sink in $T$; its capacity is the sum of capacities of directed edges from $S$ to $T$. The max-flow min-cut theorem says that maximum flow value equals minimum cut capacity. With integral capacities, there is an integral max-flow theorem attaining the maximum.

Construct a bipartite graph with one vertex for each row and column and an edge $r_i\to c_j$ whenever $A_{ij}=1$. Add source-to-row and column-to-sink edges of capacity $1$, and give row-to-column edges infinite capacity. An integral flow is precisely a set of independent $1$s, so its maximum value is the largest such set.

A finite-capacity cut places some rows on the sink side and some columns on the source side; these lines cover every $1$, since an uncovered row-to-column edge would cross the cut with infinite capacity. Its capacity is the number of selected lines. Conversely every line cover defines such a cut. Max-flow min-cut therefore proves the maximum number of independent $1$s equals the minimum number of covering lines, the matrix form of König's theorem for bipartite matching.