Dinic’s algorithm for Maximum Flow

16 May, 2022
Problem Statement : Given a graph which represents a flow network where every edge has a capacity. Also given two vertices source ‘s’ and sink ‘t’ in the graph, find the maximum possible flow from s to t with following constraints :  
Flow on an edge doesn’t exceed the given capacity of the edge.Incoming flow is equal to outgoing flow for every vertex except s and t.
Flow on an edge doesn’t exceed the given capacity of the edge.
Incoming flow is equal to outgoing flow for every vertex except s and t.
For example, in following input graph, 
the maximum s-t flow is 19 which is shown below. 
Background : 
Max Flow Problem Introduction : We introduced Maximum Flow problem, discussed Greedy Algorithm and introduced residual graph.Ford-Fulkerson Algorithm and Edmond Karp Implementation : We discussed Ford-Fulkerson algorithm and its implementation. We also discussed residual graph in detail.
Max Flow Problem Introduction : We introduced Maximum Flow problem, discussed Greedy Algorithm and introduced residual graph.
Ford-Fulkerson Algorithm and Edmond Karp Implementation : We discussed Ford-Fulkerson algorithm and its implementation. We also discussed residual graph in detail.
Time complexity of Edmond Karp Implementation is O(VE2). In this post, a new Dinic’s algorithm is discussed which is a faster algorithm and takes O(EV2).
Like Edmond Karp’s algorithm, Dinic’s algorithm uses following concepts : 
A flow is maximum if there is no s to t path in residual graph.BFS is used in a loop. There is a difference though in the way we use BFS in both algorithms.
A flow is maximum if there is no s to t path in residual graph.
BFS is used in a loop. There is a difference though in the way we use BFS in both algorithms.
In Edmond’s Karp algorithm, we use BFS to find an augmenting path and send flow across this path. In Dinic’s algorithm, we use BFS to check if more flow is possible and to construct level graph. In level graph, we assign levels to all nodes, level of a node is shortest distance (in terms of number of edges) of the node from source. Once level graph is constructed, we send multiple flows using this level graph. This is the reason it works better than Edmond Karp. In Edmond Karp, we send only flow that is send across the path found by BFS.
Outline of Dinic’s algorithm : 
1) Initialize residual graph G as given graph.
1) Do BFS of G to construct a level graph (or
   assign levels to vertices) and also check if 
   more flow is possible.
    a) If more flow is not possible, then return.
    b) Send multiple flows in G using level graph 
       until blocking flow is reached. Here using 
       level graph means, in every flow,
       levels of path nodes should be 0, 1, 2...
       (in order) from s to t.
A flow is Blocking Flow if no more flow can be sent using level graph, i.e., no more s-t path exists such that path vertices have current levels 0, 1, 2... in order. Blocking Flow can be seen same as maximum flow path in Greedy algorithm discussed here.
Illustration : Initial Residual Graph (Same as given Graph)  
Total Flow = 0
First Iteration : We assign levels to all nodes using BFS. We also check if more flow is possible (or there is a s-t path in residual graph).  
Now we find blocking flow using levels (means every flow path should have levels as 0, 1, 2, 3). We send three flows together. This is where it is optimized compared to Edmond Karp where we send one flow at a time. 4 units of flow on path s – 1 – 3 – t. 6 units of flow on path s – 1 – 4 – t. 4 units of flow on path s – 2 – 4 – t. Total flow = Total flow + 4 + 6 + 4 = 14After one iteration, residual graph changes to following.  
Second Iteration : We assign new levels to all nodes using BFS of above modified residual graph. We also check if more flow is possible (or there is a s-t path in residual graph). 
Now we find blocking flow using levels (means every flow path should have levels as 0, 1, 2, 3, 4). We can send only one flow this time. 5 units of flow on path s – 2 – 4 – 3 – t Total flow = Total flow + 5 = 19The new residual graph is  
Third Iteration : We run BFS and create a level graph. We also check if more flow is possible and proceed only if possible. This time there is no s-t path in residual graph, so we terminate the algorithm.
Implementation : Below is c++ implementation of Dinic’s algorithm:  
CPP
Python3
// C++ implementation of Dinic's Algorithm#include<bits/stdc++.h>using namespace std; // A structure to represent a edge between// two vertexstruct Edge{    int v ;  // Vertex v (or "to" vertex)             // of a directed edge u-v. "From"             // vertex u can be obtained using             // index in adjacent array.     int flow ; // flow of data in edge     int C;    // capacity     int rev ; // To store index of reverse              // edge in adjacency list so that              // we can quickly find it.}; // Residual Graphclass Graph{    int V; // number of vertex    int *level ; // stores level of a node    vector< Edge > *adj;public :    Graph(int V)    {        adj = new vector<Edge>[V];        this->V = V;        level = new int[V];    }     // add edge to the graph    void addEdge(int u, int v, int C)    {        // Forward edge : 0 flow and C capacity        Edge a{v, 0, C, adj[v].size()};         // Back edge : 0 flow and 0 capacity        Edge b{u, 0, 0, adj[u].size()};         adj[u].push_back(a);        adj[v].push_back(b); // reverse edge    }     bool BFS(int s, int t);    int sendFlow(int s, int flow, int t, int ptr[]);    int DinicMaxflow(int s, int t);}; // Finds if more flow can be sent from s to t.// Also assigns levels to nodes.bool Graph::BFS(int s, int t){    for (int i = 0 ; i < V ; i++)        level[i] = -1;     level[s] = 0;  // Level of source vertex     // Create a queue, enqueue source vertex    // and mark source vertex as visited here    // level[] array works as visited array also.    list< int > q;    q.push_back(s);     vector<Edge>::iterator i ;    while (!q.empty())    {        int u = q.front();        q.pop_front();        for (i = adj[u].begin(); i != adj[u].end(); i++)        {            Edge &e = *i;            if (level[e.v] < 0  && e.flow < e.C)            {                // Level of current vertex is,                // level of parent + 1                level[e.v] = level[u] + 1;                 q.push_back(e.v);            }        }    }     // IF we can not reach to the sink we    // return false else true    return level[t] < 0 ? false : true ;} // A DFS based function to send flow after BFS has// figured out that there is a possible flow and// constructed levels. This function called multiple// times for a single call of BFS.// flow : Current flow send by parent function call// start[] : To keep track of next edge to be explored.//           start[i] stores  count of edges explored//           from i.//  u : Current vertex//  t : Sinkint Graph::sendFlow(int u, int flow, int t, int start[]){    // Sink reached    if (u == t)        return flow;     // Traverse all adjacent edges one -by - one.    for (  ; start[u] < adj[u].size(); start[u]++)    {        // Pick next edge from adjacency list of u        Edge &e = adj[u][start[u]];                                             if (level[e.v] == level[u]+1 && e.flow < e.C)        {            // find minimum flow from u to t            int curr_flow = min(flow, e.C - e.flow);             int temp_flow = sendFlow(e.v, curr_flow, t, start);             // flow is greater than zero            if (temp_flow > 0)            {                // add flow  to current edge                e.flow += temp_flow;                 // subtract flow from reverse edge                // of current edge                adj[e.v][e.rev].flow -= temp_flow;                return temp_flow;            }        }    }     return 0;} // Returns maximum flow in graphint Graph::DinicMaxflow(int s, int t){    // Corner case    if (s == t)        return -1;     int total = 0;  // Initialize result     // Augment the flow while there is path    // from source to sink    while (BFS(s, t) == true)    {        // store how many edges are visited        // from V { 0 to V }        int *start = new int[V+1] {0};         // while flow is not zero in graph from S to D        while (int flow = sendFlow(s, INT_MAX, t, start))             // Add path flow to overall flow            total += flow;    }     // return maximum flow    return total;} // Driver Codeint main(){    Graph g(6);    g.addEdge(0, 1, 16 );    g.addEdge(0, 2, 13 );    g.addEdge(1, 2, 10 );    g.addEdge(1, 3, 12 );    g.addEdge(2, 1, 4 );    g.addEdge(2, 4, 14);    g.addEdge(3, 2, 9 );    g.addEdge(3, 5, 20 );    g.addEdge(4, 3, 7 );    g.addEdge(4, 5, 4);     // next exmp    /*g.addEdge(0, 1, 3 );      g.addEdge(0, 2, 7 ) ;      g.addEdge(1, 3, 9);      g.addEdge(1, 4, 9 );      g.addEdge(2, 1, 9 );      g.addEdge(2, 4, 9);      g.addEdge(2, 5, 4);      g.addEdge(3, 5, 3);      g.addEdge(4, 5, 7 );      g.addEdge(0, 4, 10);      // next exp     g.addEdge(0, 1, 10);     g.addEdge(0, 2, 10);     g.addEdge(1, 3, 4 );     g.addEdge(1, 4, 8 );     g.addEdge(1, 2, 2 );     g.addEdge(2, 4, 9 );     g.addEdge(3, 5, 10 );     g.addEdge(4, 3, 6 );     g.addEdge(4, 5, 10 ); */     cout << "Maximum flow " << g.DinicMaxflow(0, 5);    return 0;}
# Python implementation of Dinic's Algorithmclass Edge:    def __init__(self,v,flow,C,rev):        self.v = v        self.flow = flow        self.C = C        self.rev = rev         # Residual Graphclass Graph:    def __init__(self,V):        self.adj = [[] for i in range(V)]        self.V = V        self.level = [0 for i in range(V)]             # add edge to the graph    def addEdge(self,u,v,C):               # Forward edge : 0 flow and C capacity        a = Edge(v,0,C,len(self.adj[v]))                 # Back edge : 0 flow and 0 capacity        b = Edge(u,0,0,len(self.adj[u]))        self.adj[u].append(a)        self.adj[v].append(b)             # Finds if more flow can be sent from s to t    # Also assigns levels to nodes    def BFS(self,s,t):        for i in range(self.V):            self.level[i] = -1                     # Level of source vertex        self.level[s] = 0                 # Create a queue, enqueue source vertex        # and mark source vertex as visited here        # level[] array works as visited array also        q = []        q.append(s)        while q:            u = q.pop(0)            for i in range(len(self.adj[u])):                e = self.adj[u][i]                if self.level[e.v] < 0 and e.flow < e.C:                                       # Level of current vertex is                    # level of parent + 1                    self.level[e.v] = self.level[u]+1                    q.append(e.v)                             # If we can not reach to the sink we        # return False else True        return False if self.level[t]<0 else True       # A DFS based function to send flow after BFS has# figured out that there is a possible flow and# constructed levels. This functions called multiple# times for a single call of BFS.# flow : Current flow send by parent function call# start[] : To keep track of next edge to be explored#           start[i] stores count of edges explored#           from i# u : Current vertex# t : Sink    def sendFlow(self,u,flow,t,start):        # Sink reached        if u == t:            return flow         # Traverse all adjacent edges one -by -one        while start[u] < len(self.adj[u]):                       # Pick next edge from adjacency list of u            e = self.adj[u][start[u]]            if self.level[e.v] == self.level[u]+1 and e.flow < e.C:                               # find minimum flow from u to t                curr_flow = min(flow,e.C-e.flow)                temp_flow = self.sendFlow(e.v,curr_flow,t,start)                                 # flow is greater than zero                if temp_flow and temp_flow > 0:                                       # add flow to current edge                    e.flow += temp_flow                                         # subtract flow from reverse edge                    # of current edge                    self.adj[e.v][e.rev].flow -= temp_flow                    return temp_flow            start[u] += 1                 # Returns maximum flow in graph    def DinicMaxflow(self,s,t):               # Corner case        if s == t:            return -1                   # Initialize result        total = 0                 # Augument the flow while there is path        # from source to sink        while self.BFS(s,t) == True:                       # store how many edges are visited            # from V { 0 to V }            start = [0 for i in range(self.V+1)]            while True:                flow = self.sendFlow(s,float('inf'),t,start)                if not flow:                    break                                     # Add path flow to overall flow                total += flow                         # return maximum flow        return totalg = Graph(6)g.addEdge(0, 1, 16)g.addEdge(0, 2, 13)g.addEdge(1, 2, 10)g.addEdge(1, 3, 12)g.addEdge(2, 1, 4)g.addEdge(2, 4, 14)g.addEdge(3, 2, 9)g.addEdge(3, 5, 20)g.addEdge(4, 3, 7)g.addEdge(4, 5, 4)print("Maximum flow", g.DinicMaxflow(0,5)) # This code is contributed by rupasriachanta421.
Output: 
Maximum flow 23
Time Complexity : O(EV2). Doing a BFS to construct level graph takes O(E) time. Sending multiple more flows until a blocking flow is reached takes O(VE) time. The outer loop runs at-most O(V) time. In each iteration, we construct new level graph and find blocking flow. It can be proved that the number of levels increase at least by one in every iteration (Refer the below reference video for the proof). So the outer loop runs at most O(V) times. Therefore overall time complexity is O(EV2). 
References : https://en.wikipedia.org/wiki/Dinic’s_algorithm https://www.youtube.com/watch?v=uM06jHdIC70This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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