यह किसी दिए गए विशिष्ट मामले के लिए मेल खाने वाली समस्या को हल करने के लिए एक सी ++ प्रोग्राम है। यहां, एन पुरुषों और एन महिलाओं को दिया गया है, प्रत्येक व्यक्ति ने वरीयता के क्रम में विपरीत लिंग के सभी सदस्यों को रैंक किया है, पुरुषों और महिलाओं से एक साथ विवाह करें कि विपरीत लिंग के दो लोग नहीं हैं जो दोनों के बजाय एक दूसरे को पसंद करेंगे वर्तमान भागीदार। सभी विवाह "स्थिर" होते हैं, यदि ऐसे कोई लोग मौजूद नहीं हैं।
एल्गोरिदम
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
उदाहरण
#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;
} आउटपुट
Woman Man 4 3 5 1 6 2 7 0