# Maximum Bipartite Matching

A matching in a Bipartite Graph is a set of the edges chosen in such a way that no two edges share an endpoint. A maximum matching is a matching of maximum size (maximum number of edges). In a maximum matching, if any edge is added to it, it is no longer a matching. There can be more than one maximum matchings for a given Bipartite Graph. **Why do we care?**

There are many real world problems that can be formed as Bipartite Matching. For example, consider the following problem: *There are M job applicants and N jobs. Each applicant has a subset of jobs that he/she is interested in. Each job opening can only accept one applicant and a job applicant can be appointed for only one job. Find an assignment of jobs to applicants in such that as many applicants as possible get jobs.*

We strongly recommend to read the following post first.

Ford-Fulkerson Algorithm for Maximum Flow Problem**Maximum Bipartite Matching and Max Flow Problem** **M**aximum **B**ipartite **M**atching (**MBP**) problem can be solved by converting it into a flow network (See this video to know how did we arrive this conclusion). Following are the steps.

**1) Build a Flow Network**

There must be a source and sink in a flow network. So we add a source and add edges from source to all applicants. Similarly, add edges from all jobs to sink. The capacity of every edge is marked as 1 unit.

**2) Find the maximum flow.**

We use Ford-Fulkerson algorithm to find the maximum flow in the flow network built in step 1. The maximum flow is actually the MBP we are looking for.

**How to implement the above approach?**

Let us first define input and output forms. Input is in the form of Edmonds matrix which is a 2D array ‘bpGraph[M][N]’ with M rows (for M job applicants) and N columns (for N jobs). The value bpGraph[i][j] is 1 if i’th applicant is interested in j’th job, otherwise 0.

Output is number maximum number of people that can get jobs.

A simple way to implement this is to create a matrix that represents adjacency matrix representation of a directed graph with M+N+2 vertices. Call the fordFulkerson() for the matrix. This implementation requires O((M+N)*(M+N)) extra space.

Extra space can be reduced and code can be simplified using the fact that the graph is bipartite and capacity of every edge is either 0 or 1. The idea is to use DFS traversal to find a job for an applicant (similar to augmenting path in Ford-Fulkerson). We call bpm() for every applicant, bpm() is the DFS based function that tries all possibilities to assign a job to the applicant.

In bpm(), we one by one try all jobs that an applicant ‘u’ is interested in until we find a job, or all jobs are tried without luck. For every job we try, we do following.

If a job is not assigned to anybody, we simply assign it to the applicant and return true. If a job is assigned to somebody else say x, then we recursively check whether x can be assigned some other job. To make sure that x doesn’t get the same job again, we mark the job ‘v’ as seen before we make recursive call for x. If x can get other job, we change the applicant for job ‘v’ and return true. We use an array maxR[0..N-1] that stores the applicants assigned to different jobs.

If bmp() returns true, then it means that there is an augmenting path in flow network and 1 unit of flow is added to the result in maxBPM().

## C++

`// A C++ program to find maximal` `// Bipartite matching.` `#include <iostream>` `#include <string.h>` `using` `namespace` `std;` `// M is number of applicants` `// and N is number of jobs` `#define M 6` `#define N 6` `// A DFS based recursive function` `// that returns true if a matching` `// for vertex u is possible` `bool` `bpm(` `bool` `bpGraph[M][N], ` `int` `u,` ` ` `bool` `seen[], ` `int` `matchR[])` `{` ` ` `// Try every job one by one` ` ` `for` `(` `int` `v = 0; v < N; v++)` ` ` `{` ` ` `// If applicant u is interested in` ` ` `// job v and v is not visited` ` ` `if` `(bpGraph[u][v] && !seen[v])` ` ` `{` ` ` `// Mark v as visited` ` ` `seen[v] = ` `true` `;` ` ` `// If job 'v' is not assigned to an` ` ` `// applicant OR previously assigned` ` ` `// applicant for job v (which is matchR[v])` ` ` `// has an alternate job available.` ` ` `// Since v is marked as visited in` ` ` `// the above line, matchR[v] in the following` ` ` `// recursive call will not get job 'v' again` ` ` `if` `(matchR[v] < 0 || bpm(bpGraph, matchR[v],` ` ` `seen, matchR))` ` ` `{` ` ` `matchR[v] = u;` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `false` `;` `}` `// Returns maximum number` `// of matching from M to N` `int` `maxBPM(` `bool` `bpGraph[M][N])` `{` ` ` `// An array to keep track of the` ` ` `// applicants assigned to jobs.` ` ` `// The value of matchR[i] is the` ` ` `// applicant number assigned to job i,` ` ` `// the value -1 indicates nobody is` ` ` `// assigned.` ` ` `int` `matchR[N];` ` ` `// Initially all jobs are available` ` ` `memset` `(matchR, -1, ` `sizeof` `(matchR));` ` ` `// Count of jobs assigned to applicants` ` ` `int` `result = 0;` ` ` `for` `(` `int` `u = 0; u < M; u++)` ` ` `{` ` ` `// Mark all jobs as not seen` ` ` `// for next applicant.` ` ` `bool` `seen[N];` ` ` `memset` `(seen, 0, ` `sizeof` `(seen));` ` ` `// Find if the applicant 'u' can get a job` ` ` `if` `(bpm(bpGraph, u, seen, matchR))` ` ` `result++;` ` ` `}` ` ` `return` `result;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Let us create a bpGraph` ` ` `// shown in the above example` ` ` `bool` `bpGraph[M][N] = {{0, 1, 1, 0, 0, 0},` ` ` `{1, 0, 0, 1, 0, 0},` ` ` `{0, 0, 1, 0, 0, 0},` ` ` `{0, 0, 1, 1, 0, 0},` ` ` `{0, 0, 0, 0, 0, 0},` ` ` `{0, 0, 0, 0, 0, 1}};` ` ` `cout << ` `"Maximum number of applicants that can get job is "` ` ` `<< maxBPM(bpGraph);` ` ` `return` `0;` `}` |

## Java

`// A Java program to find maximal` `// Bipartite matching.` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `GFG` `{` ` ` `// M is number of applicants` ` ` `// and N is number of jobs` ` ` `static` `final` `int` `M = ` `6` `;` ` ` `static` `final` `int` `N = ` `6` `;` ` ` `// A DFS based recursive function that` ` ` `// returns true if a matching for` ` ` `// vertex u is possible` ` ` `boolean` `bpm(` `boolean` `bpGraph[][], ` `int` `u,` ` ` `boolean` `seen[], ` `int` `matchR[])` ` ` `{` ` ` `// Try every job one by one` ` ` `for` `(` `int` `v = ` `0` `; v < N; v++)` ` ` `{` ` ` `// If applicant u is interested` ` ` `// in job v and v is not visited` ` ` `if` `(bpGraph[u][v] && !seen[v])` ` ` `{` ` ` ` ` `// Mark v as visited` ` ` `seen[v] = ` `true` `;` ` ` `// If job 'v' is not assigned to` ` ` `// an applicant OR previously` ` ` `// assigned applicant for job v (which` ` ` `// is matchR[v]) has an alternate job available.` ` ` `// Since v is marked as visited in the` ` ` `// above line, matchR[v] in the following` ` ` `// recursive call will not get job 'v' again` ` ` `if` `(matchR[v] < ` `0` `|| bpm(bpGraph, matchR[v],` ` ` `seen, matchR))` ` ` `{` ` ` `matchR[v] = u;` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` `// Returns maximum number` ` ` `// of matching from M to N` ` ` `int` `maxBPM(` `boolean` `bpGraph[][])` ` ` `{` ` ` `// An array to keep track of the` ` ` `// applicants assigned to jobs.` ` ` `// The value of matchR[i] is the` ` ` `// applicant number assigned to job i,` ` ` `// the value -1 indicates nobody is assigned.` ` ` `int` `matchR[] = ` `new` `int` `[N];` ` ` `// Initially all jobs are available` ` ` `for` `(` `int` `i = ` `0` `; i < N; ++i)` ` ` `matchR[i] = -` `1` `;` ` ` `// Count of jobs assigned to applicants` ` ` `int` `result = ` `0` `;` ` ` `for` `(` `int` `u = ` `0` `; u < M; u++)` ` ` `{` ` ` `// Mark all jobs as not seen` ` ` `// for next applicant.` ` ` `boolean` `seen[] =` `new` `boolean` `[N] ;` ` ` `for` `(` `int` `i = ` `0` `; i < N; ++i)` ` ` `seen[i] = ` `false` `;` ` ` `// Find if the applicant 'u' can get a job` ` ` `if` `(bpm(bpGraph, u, seen, matchR))` ` ` `result++;` ` ` `}` ` ` `return` `result;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main (String[] args)` ` ` `throws` `java.lang.Exception` ` ` `{` ` ` `// Let us create a bpGraph shown` ` ` `// in the above example` ` ` `boolean` `bpGraph[][] = ` `new` `boolean` `[][]{` ` ` `{` `false` `, ` `true` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `true` `, ` `false` `, ` `false` `,` ` ` `true` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `true` `,` ` ` `true` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `true` `}};` ` ` `GFG m = ` `new` `GFG();` ` ` `System.out.println( ` `"Maximum number of applicants that can"` `+` ` ` `" get job is "` `+m.maxBPM(bpGraph));` ` ` `}` `}` |

## Python3

`# Python program to find` `# maximal Bipartite matching.` `class` `GFG:` ` ` `def` `__init__(` `self` `,graph):` ` ` ` ` `# residual graph` ` ` `self` `.graph ` `=` `graph` ` ` `self` `.ppl ` `=` `len` `(graph)` ` ` `self` `.jobs ` `=` `len` `(graph[` `0` `])` ` ` `# A DFS based recursive function` ` ` `# that returns true if a matching` ` ` `# for vertex u is possible` ` ` `def` `bpm(` `self` `, u, matchR, seen):` ` ` `# Try every job one by one` ` ` `for` `v ` `in` `range` `(` `self` `.jobs):` ` ` `# If applicant u is interested` ` ` `# in job v and v is not seen` ` ` `if` `self` `.graph[u][v] ` `and` `seen[v] ` `=` `=` `False` `:` ` ` ` ` `# Mark v as visited` ` ` `seen[v] ` `=` `True` ` ` `'''If job 'v' is not assigned to` ` ` `an applicant OR previously assigned` ` ` `applicant for job v (which is matchR[v])` ` ` `has an alternate job available.` ` ` `Since v is marked as visited in the` ` ` `above line, matchR[v] in the following` ` ` `recursive call will not get job 'v' again'''` ` ` `if` `matchR[v] ` `=` `=` `-` `1` `or` `self` `.bpm(matchR[v],` ` ` `matchR, seen):` ` ` `matchR[v] ` `=` `u` ` ` `return` `True` ` ` `return` `False` ` ` `# Returns maximum number of matching` ` ` `def` `maxBPM(` `self` `):` ` ` `'''An array to keep track of the` ` ` `applicants assigned to jobs.` ` ` `The value of matchR[i] is the` ` ` `applicant number assigned to job i,` ` ` `the value -1 indicates nobody is assigned.'''` ` ` `matchR ` `=` `[` `-` `1` `] ` `*` `self` `.jobs` ` ` ` ` `# Count of jobs assigned to applicants` ` ` `result ` `=` `0` ` ` `for` `i ` `in` `range` `(` `self` `.ppl):` ` ` ` ` `# Mark all jobs as not seen for next applicant.` ` ` `seen ` `=` `[` `False` `] ` `*` `self` `.jobs` ` ` ` ` `# Find if the applicant 'u' can get a job` ` ` `if` `self` `.bpm(i, matchR, seen):` ` ` `result ` `+` `=` `1` ` ` `return` `result` `bpGraph ` `=` `[[` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `],` ` ` `[` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `0` `],` ` ` `[` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `],` ` ` `[` `0` `, ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `],` ` ` `[` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `],` ` ` `[` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `]]` `g ` `=` `GFG(bpGraph)` `print` `(` `"Maximum number of applicants that can get job is %d "` `%` `g.maxBPM())` `# This code is contributed by Neelam Yadav` |

## C#

`// A C# program to find maximal` `// Bipartite matching.` `using` `System;` `class` `GFG` `{` ` ` `// M is number of applicants` ` ` `// and N is number of jobs` ` ` `static` `int` `M = 6;` ` ` `static` `int` `N = 6;` ` ` `// A DFS based recursive function` ` ` `// that returns true if a matching` ` ` `// for vertex u is possible` ` ` `bool` `bpm(` `bool` `[,]bpGraph, ` `int` `u,` ` ` `bool` `[]seen, ` `int` `[]matchR)` ` ` `{` ` ` `// Try every job one by one` ` ` `for` `(` `int` `v = 0; v < N; v++)` ` ` `{` ` ` `// If applicant u is interested` ` ` `// in job v and v is not visited` ` ` `if` `(bpGraph[u, v] && !seen[v])` ` ` `{` ` ` `// Mark v as visited` ` ` `seen[v] = ` `true` `;` ` ` `// If job 'v' is not assigned to` ` ` `// an applicant OR previously assigned` ` ` `// applicant for job v (which is matchR[v])` ` ` `// has an alternate job available.` ` ` `// Since v is marked as visited in the above` ` ` `// line, matchR[v] in the following recursive` ` ` `// call will not get job 'v' again` ` ` `if` `(matchR[v] < 0 || bpm(bpGraph, matchR[v],` ` ` `seen, matchR))` ` ` `{` ` ` `matchR[v] = u;` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` `// Returns maximum number of` ` ` `// matching from M to N` ` ` `int` `maxBPM(` `bool` `[,]bpGraph)` ` ` `{` ` ` `// An array to keep track of the` ` ` `// applicants assigned to jobs.` ` ` `// The value of matchR[i] is the` ` ` `// applicant number assigned to job i,` ` ` `// the value -1 indicates nobody is assigned.` ` ` `int` `[]matchR = ` `new` `int` `[N];` ` ` `// Initially all jobs are available` ` ` `for` `(` `int` `i = 0; i < N; ++i)` ` ` `matchR[i] = -1;` ` ` `// Count of jobs assigned to applicants` ` ` `int` `result = 0;` ` ` `for` `(` `int` `u = 0; u < M; u++)` ` ` `{` ` ` `// Mark all jobs as not` ` ` `// seen for next applicant.` ` ` `bool` `[]seen = ` `new` `bool` `[N] ;` ` ` `for` `(` `int` `i = 0; i < N; ++i)` ` ` `seen[i] = ` `false` `;` ` ` `// Find if the applicant` ` ` `// 'u' can get a job` ` ` `if` `(bpm(bpGraph, u, seen, matchR))` ` ` `result++;` ` ` `}` ` ` `return` `result;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `// Let us create a bpGraph shown` ` ` `// in the above example` ` ` `bool` `[,]bpGraph = ` `new` `bool` `[,]` ` ` `{{` `false` `, ` `true` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `true` `, ` `false` `, ` `false` `,` ` ` `true` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `true` `,` ` ` `true` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `false` `},` ` ` `{` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `true` `}};` ` ` `GFG m = ` `new` `GFG();` ` ` `Console.Write( ` `"Maximum number of applicants that can"` `+` ` ` `" get job is "` `+m.maxBPM(bpGraph));` ` ` `}` `}` `//This code is contributed by nitin mittal.` |

## PHP

`<?php` `// A PHP program to find maximal` `// Bipartite matching.` `// M is number of applicants` `// and N is number of jobs` `$M` `= 6;` `$N` `= 6;` `// A DFS based recursive function` `// that returns true if a matching` `// for vertex u is possible` `function` `bpm(` `$bpGraph` `, ` `$u` `, &` `$seen` `, &` `$matchR` `)` `{` ` ` `global` `$N` `;` ` ` ` ` `// Try every job one by one` ` ` `for` `(` `$v` `= 0; ` `$v` `< ` `$N` `; ` `$v` `++)` ` ` `{` ` ` `// If applicant u is interested in` ` ` `// job v and v is not visited` ` ` `if` `(` `$bpGraph` `[` `$u` `][` `$v` `] && !` `$seen` `[` `$v` `])` ` ` `{` ` ` `// Mark v as visited` ` ` `$seen` `[` `$v` `] = true;` ` ` `// If job 'v' is not assigned to an` ` ` `// applicant OR previously assigned` ` ` `// applicant for job v (which is matchR[v])` ` ` `// has an alternate job available.` ` ` `// Since v is marked as visited in` ` ` `// the above line, matchR[v] in the following` ` ` `// recursive call will not get job 'v' again` ` ` `if` `(` `$matchR` `[` `$v` `] < 0 || bpm(` `$bpGraph` `, ` `$matchR` `[` `$v` `],` ` ` `$seen` `, ` `$matchR` `))` ` ` `{` ` ` `$matchR` `[` `$v` `] = ` `$u` `;` ` ` `return` `true;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `false;` `}` `// Returns maximum number` `// of matching from M to N` `function` `maxBPM(` `$bpGraph` `)` `{` ` ` `global` `$N` `,` `$M` `;` ` ` ` ` `// An array to keep track of the` ` ` `// applicants assigned to jobs.` ` ` `// The value of matchR[i] is the` ` ` `// applicant number assigned to job i,` ` ` `// the value -1 indicates nobody is` ` ` `// assigned.` ` ` `$matchR` `= ` `array_fill` `(0, ` `$N` `, -1);` ` ` `// Initially all jobs are available` ` ` `// Count of jobs assigned to applicants` ` ` `$result` `= 0;` ` ` `for` `(` `$u` `= 0; ` `$u` `< ` `$M` `; ` `$u` `++)` ` ` `{` ` ` `// Mark all jobs as not seen` ` ` `// for next applicant.` ` ` `$seen` `=` `array_fill` `(0, ` `$N` `, false);` ` ` `// Find if the applicant 'u' can get a job` ` ` `if` `(bpm(` `$bpGraph` `, ` `$u` `, ` `$seen` `, ` `$matchR` `))` ` ` `$result` `++;` ` ` `}` ` ` `return` `$result` `;` `}` `// Driver Code` `// Let us create a bpGraph` `// shown in the above example` `$bpGraph` `= ` `array` `(` `array` `(0, 1, 1, 0, 0, 0),` ` ` `array` `(1, 0, 0, 1, 0, 0),` ` ` `array` `(0, 0, 1, 0, 0, 0),` ` ` `array` `(0, 0, 1, 1, 0, 0),` ` ` `array` `(0, 0, 0, 0, 0, 0),` ` ` `array` `(0, 0, 0, 0, 0, 1));` `echo` `"Maximum number of applicants that can get job is "` `.maxBPM(` `$bpGraph` `);` `// This code is contributed by chadan_jnu` `?>` |

## Javascript

`<script>` ` ` `// A JavaScript program to find maximal` ` ` `// Bipartite matching.` ` ` ` ` `// M is number of applicants` ` ` `// and N is number of jobs` ` ` `let M = 6;` ` ` `let N = 6;` ` ` ` ` `// A DFS based recursive function that` ` ` `// returns true if a matching for` ` ` `// vertex u is possible` ` ` `function` `bpm(bpGraph, u, seen, matchR)` ` ` `{` ` ` `// Try every job one by one` ` ` `for` `(let v = 0; v < N; v++)` ` ` `{` ` ` `// If applicant u is interested` ` ` `// in job v and v is not visited` ` ` `if` `(bpGraph[u][v] && !seen[v])` ` ` `{` ` ` ` ` `// Mark v as visited` ` ` `seen[v] = ` `true` `;` ` ` ` ` `// If job 'v' is not assigned to` ` ` `// an applicant OR previously` ` ` `// assigned applicant for job v (which` ` ` `// is matchR[v]) has an alternate job available.` ` ` `// Since v is marked as visited in the` ` ` `// above line, matchR[v] in the following` ` ` `// recursive call will not get job 'v' again` ` ` `if` `(matchR[v] < 0 || bpm(bpGraph, matchR[v],` ` ` `seen, matchR))` ` ` `{` ` ` `matchR[v] = u;` ` ` `return` `true` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// Returns maximum number` ` ` `// of matching from M to N` ` ` `function` `maxBPM(bpGraph)` ` ` `{` ` ` `// An array to keep track of the` ` ` `// applicants assigned to jobs.` ` ` `// The value of matchR[i] is the` ` ` `// applicant number assigned to job i,` ` ` `// the value -1 indicates nobody is assigned.` ` ` `let matchR = ` `new` `Array(N);` ` ` ` ` `// Initially all jobs are available` ` ` `for` `(let i = 0; i < N; ++i)` ` ` `matchR[i] = -1;` ` ` ` ` `// Count of jobs assigned to applicants` ` ` `let result = 0;` ` ` `for` `(let u = 0; u < M; u++)` ` ` `{` ` ` `// Mark all jobs as not seen` ` ` `// for next applicant.` ` ` `let seen =` `new` `Array(N);` ` ` `for` `(let i = 0; i < N; ++i)` ` ` `seen[i] = ` `false` `;` ` ` ` ` `// Find if the applicant 'u' can get a job` ` ` `if` `(bpm(bpGraph, u, seen, matchR))` ` ` `result++;` ` ` `}` ` ` `return` `result;` ` ` `}` ` ` ` ` `// Let us create a bpGraph shown` ` ` `// in the above example` ` ` `let bpGraph = [` ` ` `[` `false` `, ` `true` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `],` ` ` `[` `true` `, ` `false` `, ` `false` `,` ` ` `true` `, ` `false` `, ` `false` `],` ` ` `[` `false` `, ` `false` `, ` `true` `,` ` ` `false` `, ` `false` `, ` `false` `],` ` ` `[` `false` `, ` `false` `, ` `true` `,` ` ` `true` `, ` `false` `, ` `false` `],` ` ` `[` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `false` `],` ` ` `[` `false` `, ` `false` `, ` `false` `,` ` ` `false` `, ` `false` `, ` `true` `]];` ` ` ` ` `document.write( ` `"Maximum number of applicants that can"` `+` ` ` `" get job is "` `+ maxBPM(bpGraph));` ` ` `</script>` |

**Output: **

Maximum number of applicants that can get job is 5

You may like to see below also:

Hopcroft–Karp Algorithm for Maximum Matching | Set 1 (Introduction)

Hopcroft–Karp Algorithm for Maximum Matching | Set 2 (Implementation)**References:**

http://www.cs.cornell.edu/~wdtseng/icpc/notes/graph_part5.pdf

http://www.youtube.com/watch?v=NlQqmEXuiC8

http://en.wikipedia.org/wiki/Maximum_matching

http://www.stanford.edu/class/cs97si/08-network-flow-problems.pdf

http://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/07NetworkFlowII-2×2.pdf

http://www.ise.ncsu.edu/fangroup/or766.dir/or766_ch7.pdf

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