stable marriage problem

Initialize all men and women to free

while there exist a free man m who still has a woman w to propose to {    w = m's highest ranked such woman to whom he has not yet proposed    if w is free       (m, w) become engaged    else some pair (m', w) already exists       if w prefers m to m'          (m, w) become engaged           m' becomes free       else          (m', w) remain engaged    }

bool woman_prefer(int prefer[2*N][N],int woman,int man1,int man2){for(int i = 0; i < N; i++){//if we encounter man1 first, which indicates man1 ranks higher than man2if(prefer[woman][i] == man1)return true;if(prefer[woman][i] == man2)return false;}}void stable_marriage(int prefer[2*N][N]){//stores partner of woman, this is our output array that stores paing infor//the value of w_partner[i] indicates the partner assigned to woman N+i.Note that//the woman number between N and 2N-1,the value -1 indicates that (N+i)th woman is freeint w_partner[N];memset(w_partner,-1,sizeof(w_partner));queue<int> free_man;//it stores the no. of free manfor(int i = 0; i < N; i++)free_man.push(i);//the proposed indice of man, proposed_indice[i] means the highest indice of //woman which he has proposed to beforeint proposed_indice[N];memset(proposed_indice,-1,sizeof(proposed_indice));while(!free_man.empty()){int man = free_man.front();free_man.pop();//弹出一个free man//this indice of highest ranked woman who has not been proposedint index_propose = ++proposed_indice[man];//the woman who has not been proposed by this manint woman = prefer[man][index_propose];//if this woman is free, then they are engagedif(w_partner[woman-N] == -1){w_partner[woman-N] = man;}// if this woman has a partnerelse{int man_other = w_partner[woman-N];//if this woman prefer man to man_other,then the woman engaged //man and dumped man_otherif(woman_prefer(prefer,woman,man,man_other)){w_partner[woman-N] = man;free_man.push(man_other);}elsefree_man.push(man);}}cout << " woman man " << endl;for(int i = 0; i < N; i++)cout << " " << i+N <<"\t"<< w_partner[i] << endl;}

int main(){/*int prefer[2*N][N] ={{7, 5, 6, 4},        {5, 4, 6, 7},        {4, 5, 6, 7},        {4, 5, 6, 7},        {0, 1, 2, 3},        {0, 1, 2, 3},        {0, 1, 2, 3},        {0, 1, 2, 3}};*/int prefer[2*N][N] ={{7, 5, 6, 9,8},        {5, 8,9, 6, 7},        {9,5, 8, 6, 7},        {8, 5, 6, 7,9},{9,8,7,6,5},        {0, 1, 2, 3,4},        {0, 1,4, 2, 3},        {4,0, 1, 2, 3},        {0, 1, 4, 3,2},{4,3,2,1,0}};stable_marriage(prefer);return 0;}

分析：每次man从他的prefer list中寻找一个然后求婚，这样求婚对数最大为n^2，每次迭代消除一对所以循环的最大次数为O(n^2)。故算法肯定收敛。而一旦某个woman配对成功后就永远保持配对（但可以换配偶），所以最终肯定是n对。而且肯定是stable的，用反证法即可证明.