Description
The stable marriage problem consists of matching members of two different sets according to the member’s preferences for the other set’s members. The input for our problem consists of:
- a set M of n males;
- a set F of n females;
- for each male and female we have a list of all the members of the opposite gender in order of preference (from the most preferable to the least).
A marriage is a one-to-one mapping between males and females. A marriage is called stable, if there is no pair (m, f) such that f ∈ F prefers m ∈ M to her current partner and m prefers f over his current partner. The stable marriage A is called male-optimal if there is no other stable marriage B, where any male matches a female he prefers more than the one assigned in A.
Given preferable lists of males and females, you must find the male-optimal stable marriage.
Input
The first line gives you the number of tests. The first line of each test case contains integer n (0 < n < 27). Next line describes n male and n female names. Male name is a lowercase letter, female name is an upper-case letter. Then go n lines, that describe preferable lists for males. Next n lines describe preferable lists for females.
Output
For each test case find and print the pairs of the stable marriage, which is male-optimal. The pairs in each test case must be printed in lexicographical order of their male names as shown in sample output. Output an empty line between test cases.
Sample Input
2 3 a b c A B C a:BAC b:BAC c:ACB A:acb B:bac C:cab 3 a b c A B C a:ABC b:ABC c:BCA A:bac B:acb C:abc
Sample Output
a A b B c C a B b A c C
poj 3487 http://poj.org/problem?id=3487
稳定婚姻问题：所谓的稳定是指在当前所有的匹配中，不存在a(m,f), b(m,f),a.m更喜欢b.f而且a.f更喜欢b.m；
解决这个问题用的是模拟的方法。即男生先表白，然后女生在选取对自己表白的男生中自己最喜欢的那个。如果男生已经匹配则不可以表白了，但女生可以接受的更好的男生，这样会造成该女生原来的对象未匹配，所以该男的得继续表白了。。。
View Code
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 #define maxn 35 8 int n; 9 int mlove[maxn][maxn], flove[maxn][maxn]; 10 // 男生对女生的喜欢程度 和 女生对男生的喜欢程度 11 12 void init(){ 13 char str[100]; 14 scanf("%d", &n); 15 getchar(); gets(str); 16 for ( int i=0; i<n; ++i ) { 17 scanf("%s", str); 18 int u; 19 for ( int j=0, N=0; str[j]; ++j ){ 20 if(islower(str[j] ) ) u = str[j]-'a'; 21 else if(isupper(str[j]) ) 22 mlove[u][N++] = str[j]-'A'; 23 } 24 } 25 for ( int i=0; i<n; ++i ) { 26 scanf("%s", str); 27 int u; 28 for ( int j=0, N=0; str[j]; ++j){ 29 if(isupper(str[j]) ) u = str[j]-'A'; 30 else if(islower(str[j]) ) 31 flove[u][str[j]-'a' ] = N++; // 该男的在女生心中的位置，越小越喜欢 32 } 33 } 34 } 35 36 int pos[maxn]; // 男的已经表白了的人数 37 int fGetLove[maxn][maxn]; //女生得到表白的男生的号码 38 int mans[maxn], fans[maxn];// 保存male answer 和 female answer 39 40 void solve(){ 41 memset(pos, 0, sizeof pos); 42 memset(mans, -1, sizeof mans); //初始未配对结果是-1 43 memset(fans, -1, sizeof fans); 44 while( 1 ) { // 表白开始 45 for ( int i=0; i<n; ++i ) 46 fGetLove[i][0] = 0; // 女生得到表白的个数 47 bool ff = 1; 48 for ( int i=0; i<n; ++i ) if(pos[i] < n && mans[i] == -1){ 49 int v = mlove[i][pos[i]++ ]; 50 fGetLove[v][++fGetLove[v][0] ] = i; 51 ff = 0; 52 } 53 if( ff ) break; 54 55 // 女生开始处理表白关系; 56 for ( int i=0; i<n; ++i ) if(fGetLove[i][0] ) { 57 int minn = 30, tt; 58 for ( int j=1; j<=fGetLove[i][0]; ++j ) 59 { 60 int v = fGetLove[i][j]; 61 if(minn > flove[i][v] ) 62 minn = flove[i][v], tt = v; 63 } 64 // 当前女生已经有了匹配对象，所以minn要和该对象的映射比较 65 if(~fans[i] && minn > flove[i][fans[i] ]) continue; 66 if( ~fans[i] ) mans[fans[i] ] = -1; // 67 fans[i] = tt; mans[tt] = i; 68 } 69 } 70 for ( int i=0; i<n; ++i ) 71 printf("%c %c\n", i+'a', mans[i]+'A'); 72 } 73 74 int main(int argc, char**argv){ 75 int T; scanf("%d", &T); 76 while( T-- ) { 77 init(); 78 solve(); 79 if( T ) puts(""); 80 } 81 //system("pause"); 82 return (EXIT_SUCCESS); 83 }
同样的问题有：
hdoj 1522 http://acm.hdu.edu.cn/showproblem.php?pid=1522
codechef http://www.codechef.com/AUG12/problems/BLOCKING
