POJ3487[稳定婚姻]

Posted 2020-09-04 神犇(shenben)

The Stable Marriage Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2974		Accepted: 1267

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

Source

Southeastern Europe 2007

题目大意：
    有N位女士和N位男士，每位男士或者女士都对对方有个评价。
他们会形成N对夫妻，如果A和a结婚，B和b结婚，但是A更偏爱b而非a而且b也更偏爱A而非B，那么这种婚姻是不稳定的 .
求 一个稳定的 婚姻 配对。

    题目要求是男士优先，也就是男士优先表白，肯定会从最喜欢的开始表白，如果对方没有男友或者表白的男士优于当前男友，就会抛弃原来的男友，而接受表白的男士，男士也成功脱光。否则男士会被拒绝，便只能考虑下一个喜欢的女士。直到所有人都脱光，不存在拒绝情况为止。
采用 Gale-Shapley算法 :
    男生不停的求婚，女生不停地拒绝.
    算法保证：每一位男性得到的伴侣都是所有可能的稳定婚姻搭配方案中最理想的，同时每一位女性得到的伴侣都是所有可能的稳定婚姻搭配方案中最差的。
是不是只有唯一的稳定婚姻呢，觉得如果没有对两个人的评价完全相同的情况下，应该是唯一的.

　

• Source Code

  #include<cstdio>
  #include<cstring>
  #include<queue>
  #include<algorithm>
  using namespace std;
  const int N=30;
  int malelike[N][N],femalelike[N][N];
  int malechoice[N],femalechoice[N];
  int malename[N],femalename[N];
  int T,couple;char str[N];
  queue<int>freemale;
  int main(){
  	scanf("%d",&T);
  	while(T--){
  		while(!freemale.empty()) freemale.pop();
  		scanf("%d",&couple);
  		for(int i=0;i<couple;i++) scanf("%s",str),freemale.push(malename[i]=str[0]-‘a‘);
  		sort(malename,malename+couple);
  		for(int i=0;i<couple;i++) scanf("%s",str),femalename[i]=str[0]-‘A‘;
  		for(int i=0;i<couple;i++){
  			scanf("%s",str);
  			for(int j=0;j<couple;j++){
  				malelike[i][j]=str[j+2]-‘A‘;
  			}
  		}
  		for(int i=0;i<couple;i++){
  			scanf("%s",str);
  			for(int j=0;j<couple;j++){
  				femalelike[i][str[j+2]-‘a‘]=couple-j;
  			}
  			femalelike[i][couple]=0;
  		}
  		memset(malechoice,0,(couple+2)<<2);
  		for(int i=0;i<couple;i++) femalechoice[i]=couple;
  		while(!freemale.empty()){
  			int male=freemale.front();
  			int female=malelike[male][malechoice[male]];
  			if(femalelike[female][male]>femalelike[female][femalechoice[female]]){
  				freemale.pop();
  				if(femalechoice[female]!=couple){
  					freemale.push(femalechoice[female]);
  					malechoice[femalechoice[female]]++;
  				}
  				femalechoice[female]=male;
  			}
  			else malechoice[male]++;
  		}
  		for(int i=0;i<couple;i++) printf("%c %c\n",malename[i]+‘a‘,malelike[malename[i]][malechoice[malename[i]]]+‘A‘);
  		if(T) putchar(‘\n‘);
  	}
  	return 0;
  }