Minimum Cost Maximum Flow from a Graph using Bellman Ford Algorithm

Given a source node S, a sink node T, two matrices Cap[ ][ ] and Cost[ ][ ] representing a graph, where Cap[i][j] is the capacity of a directed edge from node i to node j and cost[i][j] is the cost of sending one unit of flow along a directed edge from node i to node j, the task is to find a flow with the minimum-cost maximum-flow possible from the given graph.

Minimum Cost Maximum Flow: Minimum cost(per unit of flow) required to deliver maximum amount of flow possible from the given graph.

Example:

Input: S = 0, T = 4, cap[ ][ ] = {{0, 3, 4, 5, 0}, {0, 0, 2, 0, 0}, {0, 0, 0, 4, 1}, {0, 0, 0, 0, 10}, {0, 0, 0, 0, 0}},
cost[ ][ ] = {{0, 1, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}}
Output: 10 1
Explanation:
For given graph, Max flow = 10 and Min cost = 1.

Input: S = 0, T = 4, cost[][] = { { 0, 1, 0, 0, 2 }, { 0, 0, 0, 3, 0 }, { 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 1 }, { 0, 0, 0, 0, 0 } }



cap[][] = { { 0, 3, 1, 0, 3 }, { 0, 0, 2, 0, 0 }, { 0, 0, 0, 1, 6 }, { 0, 0, 0, 0, 2 }, { 0, 0, 0, 0, 0 } }
Output: 6 8

Approach:
Negative cycle in the cost network is cycled with the sum of costs of all the edges in the cycle is negative. They can be detected using Bellman Ford algorithm. They should be eliminated because, practically, flow through such cycles cannot be allowed. Consider a negative cost cycle, if all flow has to pass through this cycle, the total cost is always reducing for every cycle completed. This would result in an infinite loop in the desire of minimizing the total cost. So, whenever a cost network includes a negative cycle, it implies, the cost can further be minimized (by flowing through the other side of the cycle instead of the side currently considered). A negative cycle once detected are removed by flowing a Bottleneck Capacity through all the edges in the cycle.

Now, look at what supply and demand nodes are:

Supply nodes: These are positive Nodes that are added to the flow and which produces the flow.
Demand nodes: These are negative nodes which are subtracted from the flow.

Supply (or demand) at each node = Total flow leading out of the Node – The total flow leading into the Node

The given problem is approached by sending a bottleneck capacity to all edges in the cycle to take care of the negative cycle. Additionally, since it involves demand nodes, the Bellman Ford algorithm is invoked.
Follow the steps below to solve the problem:

  • Store the capacity of an edge and the cost of that edge in two seperate array.
  • Given the source node S and sink node T, picked edge pi, demand nodes da and distance between nodes dist, search if it’s possible to have flow from S to T.
  • If a flow exists, calculate the distance, value = dist + pi – pi[k] – cost[k].
  • Compare the distance values in dist[ ] with value and keep updating until minimum flow is obtained.

Below is the implementation of the above approach:

Java

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// Java Program to implement
// the above approach
import java.util.*;
  
public class MinCostMaxFlow {
  
    // Stores the found edges
    boolean found[];
  
    // Stores the number of nodes
    int N;
  
    // Stores the capacity
    // of each edge
    int cap[][];
  
    int flow[][];
  
    // Stores the cost per
    // unit flow of each edge
    int cost[][];
  
    // Stores the distance from each node
    // and picked edges for each node
    int dad[], dist[], pi[];
  
    static final int INF
        = Integer.MAX_VALUE / 2 - 1;
  
    // Function to check if it is possible to
    // have a flow from the src to sink
    boolean search(int src, int sink)
    {
  
        // Initialise found[] to false
        Arrays.fill(found, false);
  
        // Initialise the dist[] to INF
        Arrays.fill(dist, INF);
  
        // Distance from the source node
        dist[src] = 0;
  
        // Iterate untill src reaches N
        while (src != N) {
  
            int best = N;
            found[src] = true;
  
            for (int k = 0; k < N; k++) {
  
                // If already found
                if (found[k])
                    continue;
  
                // Evaluate while flow
                // is still in supply
                if (flow[k][src] != 0) {
  
                    // Obtain the total value
                    int val
                        = dist[src] + pi[src]
                          - pi[k] - cost[k][src];
  
                    // If dist[k] is > minimum value
                    if (dist[k] > val) {
  
                        // Update
                        dist[k] = val;
                        dad[k] = src;
                    }
                }
  
                if (flow[src][k] < cap[src][k]) {
  
                    int val = dist[src] + pi[src]
                              - pi[k] + cost[src][k];
  
                    // If dist[k] is > minimum value
                    if (dist[k] > val) {
  
                        // Update
                        dist[k] = val;
                        dad[k] = src;
                    }
                }
  
                if (dist[k] < dist[best])
                    best = k;
            }
  
            // Update src to best for
            // next iteration
            src = best;
        }
  
        for (int k = 0; k < N; k++)
            pi[k]
                = Math.min(pi[k] + dist[k],
                           INF);
  
        // Return the value obtained at sink
        return found[sink];
    }
  
    // Function to obtain the maximum Flow
    int[] getMaxFlow(int cap[][], int cost[][],
                     int src, int sink)
    {
  
        this.cap = cap;
        this.cost = cost;
  
        N = cap.length;
        found = new boolean[N];
        flow = new int[N][N];
        dist = new int[N + 1];
        dad = new int[N];
        pi = new int[N];
  
        int totflow = 0, totcost = 0;
  
        // If a path exist from src to sink
        while (search(src, sink)) {
  
            // Set the default amount
            int amt = INF;
            for (int x = sink; x != src; x = dad[x])
  
                amt = Math.min(amt,
                               flow[x][dad[x]] != 0
                                   ? flow[x][dad[x]]
                                   : cap[dad[x]][x]
                                         - flow[dad[x]][x]);
  
            for (int x = sink; x != src; x = dad[x]) {
  
                if (flow[x][dad[x]] != 0) {
                    flow[x][dad[x]] -= amt;
                    totcost -= amt * cost[x][dad[x]];
                }
                else {
                    flow[dad[x]][x] += amt;
                    totcost += amt * cost[dad[x]][x];
                }
            }
            totflow += amt;
        }
  
        // Return pair total cost and sink
        return new int[] { totflow, totcost };
    }
  
    // Driver Code
    public static void main(String args[])
    {
  
        // Creating an object flow
        MinCostMaxFlow flow = new MinCostMaxFlow();
  
        int s = 0, t = 4;
  
        int cap[][] = { { 0, 3, 1, 0, 3 },
                        { 0, 0, 2, 0, 0 },
                        { 0, 0, 0, 1, 6 },
                        { 0, 0, 0, 0, 2 },
                        { 0, 0, 0, 0, 0 } };
  
        int cost[][] = { { 0, 1, 0, 0, 2 },
                         { 0, 0, 0, 3, 0 },
                         { 0, 0, 0, 0, 0 },
                         { 0, 0, 0, 0, 1 },
                         { 0, 0, 0, 0, 0 } };
  
        int ret[] = flow.getMaxFlow(cap, cost, s, t);
  
        System.out.println(ret[0] + " " + ret[1]);
    }
}

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Output:

6 8

Time Complexity: O(V2 * E2) where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V)

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