Pku3487 (marital stability issues)

Http: // 162.105.81.212/judgeonline/problem? Id = 3487

If you don't understand the meaning of the question, read my previous blog. The following describes the related algorithms: refer to the algorithm of Niu Ren;

 

Gale-Shapley Algorithm

While there is a man m who is free and has not been married to every woman

Choose this man m
Make w the highest rank of Women in the priority list of M who have not yet been married
If W is free
(M, W) Changes to appointment status
Else W is currently dating M1
If W prefers M1 rather than m
M keep free
Else W prefers m rather than M1
(M, W) Changes to appointment status
M1 becomes free
Endif
Endif
Endwhile

# Include <iostream> <br/> # include <queue> <br/> using namespace STD; <br/> int MAN [30] [30], WOMAN [30] [30]; <br/> int guple [30], MMP re [30]; <br/> int A [30], N; <br/> queue <int> q; <br/> void gale_shapley () <br/> {<br/> memset (guple, 0, sizeof (guple )); <br/> memset (matrix,-1, sizeof (matrix); <br/> int I, PM, PF; <br/> while (! Q. empty () <br/>{< br/> PM = Q. front (); <br/> q. pop (); <br/> pF = WOMAN [PM] [guple [PM]; <br/> guple [PM] ++; <br/> If (MMP re [pf] <0) // PF is free (PM, PF) and changes to appointment status <br/> MMP re [pf] = PM; <br/> else if (MAN [pf] [MMP re [pf] <MAN [pf] [PM]) // pf prefers PM1, PM keep free <br/> q. push (PM); <br/> else // pf prefers pm, instead of PM1. (PM, Pf) Changes to appointment status; <br/>{< br/> q. push (Matrix [pf]); <br/> matrix [pf] = PM; <br/>}< br/> for (I = 0; I <26; I ++) <br/> If (MMP re [I] >-1) guple [matrix [I] = I; <br/> for (I = 0; I <n; I ++) <br/> printf ("% C/N", a [I] + 'A', guple [A [I] + 'A '); <br/> puts (""); <br/>}< br/> int main () <br/>{< br/> int T, I, J; <br/> char CH, STR [30]; <br/> scanf ("% d", & T); <br/> while (t --) <br/>{< br/> scanf ("% d", & N); <br/> while (! Q. empty () <br/> q. pop (); <br/> for (I = 0; I <n; I ++) <br/> {<br/> scanf ("% C ", & Ch); // note that there is a space reserved before it, which is repeated in scanf ("% C ..) <br/> A [I] = CH-'A'; <br/> q. push (CH-'A'); <br/>}< br/> sort (A, A + n); <br/> for (I = 0; I <n; I ++) <br/> scanf ("% C", & Ch); <br/> for (I = 0; I <N; I ++) <br/>{< br/> scanf ("% s", STR); <br/> for (j = 0; j <N; j ++) <br/> WOMAN [STR [0]-'a'] [J] = STR [J + 2]-'A '; <br/>}< br/> for (I = 0; I <n; I ++) <br/> {<br/> scanf ("% s ", str); <br/> for (j = 0; j <n; j ++) <br/> MAN [STR [0]-'a'] [STR [J + 2]-'a'] = J; // pay attention to the serial number; <br/>}</P> <p> gale_shapley (); <br/> // for (I = 0; I <n; I ++) <br/> // printf ("% C/N", a [I] + 'A', womanpre [A [I] + 'A '); <br/>}< br/> return 0; <br/>}< br/>