C++ Program to Solve a Matching Problem for a Given Specific Case

This is a C++ Program to solve a matching problem for a Given Specific Case. Here, N men and N women are given, each person has ranked all members of the opposite gender in order of preference, marry the men and women together such that there are no two people of opposite gender who would both rather have each other than their current partners. All the marriages are “stable”, if there are no such people exists.

Algorithms

Begin
   function WomenPrefersMenOverMen1():
   A) Check if women prefer men over her current engagement men1
   B) If men1 comes before men in list of women, then women prefer her current engagement.
   C) If men comes before men1 in womens's list, then free her current engagement and engage her with men.
End
Begin
   function stablewedding():
   1) Boys are numbered as 0 to N-1.
   2) Girls are numbered as N to 2N-1.
   3) While men are free
      A) Pick the first free man
      B) One by one go to all women according to pick free man’s preferences.
      C) The woman of preference is free, woman and man become partners.
      D) If woman is not free Find current engagement of woman
      E) If woman prefers man over her current engagement man1, then break the engagement between woman and man1 and engage man with woman.
End

Example

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define N 4
bool WomenPrefersMenOverMen1(int prefer[2*N][N], int w, int m, int m1) {
   for (int i = 0; i < N; i++) {
      if (prefer[w][i] == m1)
         return true;
      if (prefer[w][i] == m)
         return false;
   }
}
void stablewedding(int prefer[2*N][N]) {
   int wPartner[N]; //Initialize an array to store partner of women.
   bool mFree[N]; //Initialize an array to store availability of men.
   //Initialize all men and women as free.
   memset(wPartner, -1, sizeof(wPartner));
   memset(mFree, false, sizeof(mFree));
   int freeCnt = N;
   while (freeCnt > 0) { //While men is free
      int m; //Pick the first free man
      //One by one go to all women according to pick free man’s preferences.
      for (m = 0; m < N; m++)
         if (mFree[m] == false)
            break;
      for (int i = 0; i < N && mFree[m] == false; i++) {
         int w = prefer[m][i];
         //The woman of preference is free, woman and man become partners.
         if (wPartner[w-N] == -1) {
            wPartner[w-N] = m;
            mFree[m] = true;
            freeCnt--;
         } else { //If w is not free
             //Find current engagement of woman
            int m1 = wPartner[w-N];
            // If woman prefers man over her current engagement man1, 
            // then break the engagement between woman and man1 and engage man with woman.
            if (WomenPrefersMenOverMen1(prefer, w, m, m1) == false) {
               wPartner[w-N] = m;
               mFree[m] = true;
               mFree[m1] = false;
            }
         }      
      }
   }
   cout << "Woman Man" << endl;
   for (int i = 0; i < N; i++)
      cout << " " << i+N << "\t" << wPartner[i] << endl;
}
int main() {
   int p[2*N][N] = { 
      {7, 5, 6, 4},
      {5, 4, 7, 6},
      {4, 5, 7, 6},
      {4, 5, 7, 6},
      {0, 1, 3, 2},
      {0, 1, 3, 2},
      {0, 1, 3, 2},
      {0, 1, 3, 2},
   };
   stablewedding(p);
   return 0;
}

Output

Woman Man
4 3
5 1
6 2
7 0

Smita Kapse

Published on 03-May-2019 11:36:57

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