Ford Fulkerson

FordFulkerson.java

int fordFulkerson(int n, int s, int t) {
        //ASSUMES: cap[u][v] stores capacity of edge (u,v). cap[u][v] = 0 for no edge.
        //Initialise the flow network so that fnet[u][v] = 0 for all u,v
        int flow = 0; //no flow yet

        while (true) {
            //Find an augmenting path using BFS
            int[] prev = new int[n];
            Arrays.fill(prev, -1);
            LinkedList<Integer> queue = new LinkedList<Integer>();
            prev[s] = -2;
            queue.add(s);
            while (!queue.isEmpty() && prev[t] == -1) {
                int u = queue.poll();
                for (int v = 0; v < n; v++) {
                    if (prev[v] == -1) { //not seen yet
                        if (fnet[v][u] > 0 || fnet[u][v] < cap[u][v]) {
                            prev[v] = u;
                            queue.add(v);
                        }
                    }
                }
            }
            //See if we couldn't find any path to t (t has no parents)
            if (prev[t] == -1) break;

            //Get the bottleneck capacity;
            int bot = Integer.MAX_VALUE;
            for (int v = t, u = prev[v]; u >= 0; v = u, u = prev[v]) {
                if (fnet[v][u] > 0)  //prefer a backward edge over a forward
                    bot = min(bot, fnet[v][u]);
                else //must be a forward edge otherwise
                    bot = min(bot, cap[u][v] - fnet[u][v]);
            }

            //update the flow network
            for (int v = t, u = prev[v]; u >= 0; v = u, u = prev[v]) {
                if (fnet[v][u] > 0) //backward edge -> subtract
                    fnet[v][u] -= bot;
                else //forward edge -> add
                    fnet[u][v] += bot;
            }

            //Sent 'bot' amount of flow from s to v, so update the flow
            flow += bot;
        }
        return flow;
    }

This is actually Edmonds-Karp algorithm since BFS is used to find the augment path.

Looks good except it would be nice if you labelled n, s, and t.