The Stable Marriage Problem

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3274 Accepted: 1401

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 (mf) 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

`#include<cstdio>#include<queue>#include<algorithm>#include<cstring>#include<iostream>using namespace std;const int maxn=30;int couple;//总共多少对int malelike[maxn][maxn],femalelike[maxn][maxn];//男士对女士的喜欢程度（按降序排列）和女士对男士的喜欢程度 int malechoice[maxn],femalechoice[maxn];//男士和女士的选择,男士选择了第几喜欢的 int malename[maxn],femalename[maxn];//名字的hash,方便打印对应编号的名字 int main(){  int T;  char str[30];  scanf("%d",&T);  while(T--)  {    queue<int> freemale;//没有配对的男士     scanf("%d",&couple);    for(int i=0;i<couple;i++)    {      scanf("%s",str);      malename[i]=str[0]-'a';      freemale.push(malename[i]);    }    //将名字排序，便于字典序    sort(malename,malename+couple);    for(int i=0;i<couple;i++)    {      //女士是大写       scanf("%s",str);      femalename[i]=str[0]-'A';    }    //男士对女士的印象，按降序排列    for(int i=0;i<couple;i++)    {      scanf("%s",str);      for(int j=0;j<couple;j++)      malelike[i][j]=str[j+2]-'A';//他喜欢的是谁     }    //女士对男士的打分，添加虚拟人物，编号为couple，为女士的初始对象    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;//她喜欢他多少       femalelike[i][couple]=0;    }    //初始化男士选自己最喜欢的女士，其实还是光棍     memset(malechoice,0,sizeof(malechoice));    //女士先初始一个对象    for(int i=0;i<couple;i++)    femalechoice[i]=couple;    while(!freemale.empty())    {      //找出一个未配对的男士，注意不要习惯性的pop      int male=freemale.front();      //男士心仪的女士      int female=malelike[male][malechoice[male]];      //如果当前的男士比原来的男友更好      if(femalelike[female][male]>femalelike[female][femalechoice[female]])      {        //该男士成功脱单        freemale.pop();        //如果有前男友，则把前男友打回光棍,则该光棍只能考虑下一个女士咯         //不要把虚拟的人物加入队列，否则死循环或者错误         if(femalechoice[female]!=couple)        {          freemale.push(femalechoice[female]);          malechoice[femalechoice[female]]++;        }        //当前男友为这位男士        femalechoice[female]=male;       }      //如果被该女士拒接，则只能找下一个女士咯      else      malechoice[male]++;    }    for(int i=0;i<couple;i++)    printf("%c %c\n",malename[i]+'a',    malelike[malename[i]][malechoice[malename[i]]]+'A');    puts("");  }  return 0;}`