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 separate 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:
C++
#include <iostream> #include <vector> #include <climits> using namespace std; // Stores the found edges vector< bool > found; // Stores the number of nodes int N = 0; // Stores the capacity of each edge vector<vector< int >> cap; vector<vector< int >> flow; // Stores the cost per unit flow of each edge vector<vector< int >> cost; // Stores the distance from each node // and picked edges for each node vector< int > dad, dist, pi; const int INF = INT_MAX / 2 - 1; // Function to check if it is possible to // have a flow from the src to sink bool search( int src, int sink) { // Initialise found[] to false found = vector< bool >(N, false ); // Initialise the dist[] to INF dist = vector< int >(N + 1, INF); // Distance from the source node dist[src] = 0; // Iterate until 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] = min(pi[k] + dist[k], INF); // Return the value obtained at sink return found[sink]; } // Function to obtain the maximum Flow vector< int > getMaxFlow(vector<vector< int >>& capi, vector<vector< int >>& costi, int src, int sink) { cap = capi; cost = costi; N = cap.size(); found = vector< bool >(N, false ); flow.assign(N, vector< int >(N, 0)); dist = vector< int >(N + 1, 0); dad = vector< int >(N, 0); pi = vector< int >(N, 0); int totflow = 0, totcost = 0; // If a path exists from src to sink while (search(src, sink)) { // Set the default amount int amt = INF; int x = sink; while (x != src) { amt = min( amt, flow[x][dad[x]] != 0 ? flow[x][dad[x]] : cap[dad[x]][x] - flow[dad[x]][x]); x = dad[x]; } x = sink; while (x != src) { 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]; } x = dad[x]; } totflow += amt; } // Return pair total cost and sink return {totflow, totcost}; } // Driver Code int main() { int s = 0, t = 4; vector<vector< 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}}; vector<vector< 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}}; vector< int > ret = getMaxFlow(cap, cost, s, t); cout << ret[0] << " " << ret[1] << endl; return 0; } // by phasing17 |
Java
// 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 until 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 ]); } } |
Python3
# Python3 program to implement # the above approach from sys import maxsize from typing import List # Stores the found edges found = [] # Stores the number of nodes N = 0 # Stores the capacity # of each edge cap = [] flow = [] # Stores the cost per # unit flow of each edge cost = [] # Stores the distance from each node # and picked edges for each node dad = [] dist = [] pi = [] INF = maxsize / / 2 - 1 # Function to check if it is possible to # have a flow from the src to sink def search(src: int , sink: int ) - > bool : # Initialise found[] to false found = [ False for _ in range (N)] # Initialise the dist[] to INF dist = [INF for _ in range (N + 1 )] # Distance from the source node dist[src] = 0 # Iterate until src reaches N while (src ! = N): best = N found[src] = True for k in range (N): # If already found if (found[k]): continue # Evaluate while flow # is still in supply if (flow[k][src] ! = 0 ): # Obtain the total value 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]): 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 k in range (N): pi[k] = min (pi[k] + dist[k], INF) # Return the value obtained at sink return found[sink] # Function to obtain the maximum Flow def getMaxFlow(capi: List [ List [ int ]], costi: List [ List [ int ]], src: int , sink: int ) - > List [ int ]: global cap, cost, found, dist, pi, N, flow, dad cap = capi cost = costi N = len (capi) found = [ False for _ in range (N)] flow = [[ 0 for _ in range (N)] for _ in range (N)] dist = [INF for _ in range (N + 1 )] dad = [ 0 for _ in range (N)] pi = [ 0 for _ in range (N)] totflow = 0 totcost = 0 # If a path exist from src to sink while (search(src, sink)): # Set the default amount amt = INF x = sink while x ! = src: amt = min ( amt, flow[x][dad[x]] if (flow[x][dad[x]] ! = 0 ) else cap[dad[x]][x] - flow[dad[x]][x]) x = dad[x] x = sink while x ! = src: 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] x = dad[x] totflow + = amt # Return pair total cost and sink return [totflow, totcost] # Driver Code if __name__ = = "__main__" : s = 0 t = 4 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 ] ] 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 ] ] ret = getMaxFlow(cap, cost, s, t) print ( "{} {}" . format (ret[ 0 ], ret[ 1 ])) # This code is contributed by sanjeev2552 |
C#
// C# Program to implement // the above approach using System; using System.Collections.Generic; public class MinCostMaxFlow { // Stores the found edges bool [] 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 int INF = Int32.MaxValue / 2 - 1; // Function to check if it is possible to // have a flow from the src to sink bool search( int src, int sink) { // Initialise found[] to false for ( int i = 0; i < found.Length; i++) found[i] = false ; // Initialise the dist[] to INF for ( int i = 0; i < dist.Length; i++) dist[i] = INF; // Distance from the source node dist[src] = 0; // Iterate until 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.GetLength(0); found = new bool [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); Console.WriteLine(ret[0] + " " + ret[1]); } } // This code is contributed by phasing17. |
Javascript
// JS program to implement // the above approach let maxsize = Number.MAX_VALUE // Stores the found edges let found = [] // Stores the number of nodes let N = 0 // Stores the capacity // of each edge let cap = [] let flow = [] // Stores the cost per // unit flow of each edge let cost = [] // Stores the distance from each node // and picked edges for each node let dad = [] let dist = [] let pi = [] let INF = Math.floor(maxsize / 2) - 1 // Function to check if it is possible to // have a flow from the src to sink function search(src, sink) { // Initialise found[] to false let found = new Array(N).fill( false ) // Initialise the dist[] to INF let dist = new Array(N + 1).fill(INF) // Distance from the source node dist[src] = 0 // Iterate until src reaches N while (src != N) { let best = N found[src] = true for ( var 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 let 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]) { let 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 ( var 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 function getMaxFlow(capi, costi, src, sink) { cap = capi cost = costi N = (capi).length found = new Array(N).fill( false ); flow = new Array(N); for ( var i = 0; i < N; i++) flow[i] = new Array(N).fill(0) dist = new Array(N + 1).fill(INF) dad = new Array(N).fill(0) pi = new Array(N).fill(0) totflow = 0 totcost = 0 // If a path exist from src to sink while (search(src, sink)) { // Set the default amount amt = INF x = sink while (x != src) { amt = Math.min( amt, (flow[x][dad[x]] != 0)?flow[x][dad[x]]: cap[dad[x]][x] - flow[dad[x]][x]) x = dad[x] } x = sink while (x != src) { 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] } x = dad[x] } totflow += amt } // Return pair total cost and sink return [totflow, totcost] } // Driver Code let s = 0 let t = 4 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 ] ] 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 ] ] let ret = getMaxFlow(cap, cost, s, t) console.log(ret[0], ret[1]) // This code is contributed by phasing17 |
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)