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.
题目大意：就是稳定婚姻问题，要求男士最优
思路：直接套用Gale-Shapley算法即可
PS：直接用数字不就好了吗非要用字符……
1 #include <cstdio> 2 #include <queue> 3 #include <cstring> 4 #include <map> 5 using namespace std; 6 7 const int MAXN = 100; 8 9 int pref[MAXN][MAXN], order[MAXN][MAXN], next[MAXN]; 10 int future_husband[MAXN], future_wife[MAXN]; 11 queue<int> que; 12 map<char, int> mp; 13 14 void engage(int man, int woman){ 15 int &m = future_husband[woman]; 16 if(m){ 17 future_wife[m] = 0; 18 que.push(m); 19 } 20 future_husband[woman] = man; 21 future_wife[man] = woman; 22 } 23 24 int n, T; 25 26 void GaleShapley(){ 27 while(!que.empty()){ 28 int man = que.front(); que.pop(); 29 int woman = pref[man][next[man]++]; 30 if(!future_husband[woman] || order[woman][man] < order[woman][future_husband[woman]]) 31 engage(man, woman); 32 else que.push(man); 33 } 34 for(char c = 'a'; c <= 'z'; ++c) if(mp[c]) 35 printf("%c %c ", c, future_wife[mp[c]] + 'A' - 1); 36 } 37 38 int main(){ 39 char s[MAXN], c[2]; 40 scanf("%d", &T); 41 while(T--){ 42 if(!que.empty()) que.pop(); 43 mp.clear(); 44 memset(pref,0,sizeof(pref)); 45 memset(order,0,sizeof(order)); 46 memset(future_husband,0,sizeof(future_husband)); 47 memset(future_wife,0,sizeof(future_wife)); 48 scanf("%d", &n); 49 for(int i = 1; i <= n; ++i) scanf("%s", c), mp[c[0]] = i; 50 for(int i = 1; i <= n; ++i) scanf("%s", c), mp[c[0]] = i; 51 for(int i = 0; i < n; ++i){ 52 scanf("%s", s); 53 for(int j = 2; s[j]; ++j) pref[mp[s[0]]][j-1] = mp[s[j]]; 54 next[mp[s[0]]] = 1; 55 que.push(mp[s[0]]); 56 } 57 for(int i = 0; i < n; ++i){ 58 scanf("%s", s); 59 for(int j = 2; s[j]; ++j) order[mp[s[0]]][mp[s[j]]] = j-1; 60 } 61 GaleShapley(); 62 if(T) printf(" "); 63 } 64 }