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HDOJ 1914 The Stable Marriage Problem

2024-07-25 05:03:31   220人阅读


rt 稳定婚姻匹配问题

The Stable Marriage Problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 438    Accepted Submission(s): 222


Problem 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
 



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

int n;
char boy_name[30][2],girl_name[30][2];
int to_boy[26],to_girl[26];

int perfect_boy[30][30],perfect_girl[30][30];
int future_husband[30],future_wife[30];
int next[30];
queue<int> q;

void init()
{
    memset(boy_name,0,sizeof(boy_name));
    memset(girl_name,0,sizeof(girl_name));
    memset(to_boy,0,sizeof(to_boy));
    memset(to_girl,0,sizeof(to_girl));
    memset(perfect_boy,0,sizeof(perfect_boy));
    memset(perfect_girl,0,sizeof(perfect_girl));
    memset(future_husband,0,sizeof(future_husband));
    memset(future_wife,0,sizeof(future_wife));
    memset(next,0,sizeof(next));
    while(!q.empty()) q.pop();
}

void engage(int boy,int girl)
{
    int m=future_husband[girl];
    if(m)
    {
        future_wife[m]=0;
        q.push(m);
    }
    future_husband[girl]=boy;
    future_wife[boy]=girl;
}

bool lover(int boy,int m,int girl)
{
    for(int i=1;i<=n;i++)
    {
        if(perfect_boy[girl][i]==boy) return true;
        if(perfect_boy[girl][i]==m) return false;
    }
}

int main()
{
    int T_T,flag=0;
    char in[50];
    scanf("%d",&T_T);
    while(T_T--)
    {
        if(flag==0) flag=1;
        else putchar(10);
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",boy_name[i]);
            to_boy[boy_name[i][0]-'a']=i;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",girl_name[i]);
            to_girl[girl_name[i][0]-'A']=i;
        }
        for(int i=0;i<n;i++)
        {
            scanf("%s",in);
            int boy=to_boy[in[0]-'a'];
            for(int j=2;j<n+2;j++)
            {
                int girl=to_girl[in[j]-'A'];
                perfect_girl[boy][j-1]=girl;
            }
            q.push(i+1);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%s",in);
            int girl=to_girl[in[0]-'A'];
            for(int j=2;j<n+2;j++)
            {
                int boy=to_boy[in[j]-'a'];
                perfect_boy[girl][j-1]=boy;
            }
        }
        while(!q.empty())
        {
            int boy=q.front(); q.pop();
            int girl=perfect_girl[boy][++next[boy]];
            int m=future_husband[girl];
            if(m==0)
                engage(boy,girl);
            else
            {
                if(lover(boy,m,girl))
                    engage(boy,girl);
                else q.push(boy);
            }
        }
        for(int i=1;i<=n;i++)
        {
            int boy=to_boy[boy_name[i][0]-'a'];
            printf("%c %c\n",boy_name[i][0],girl_name[future_wife[boy]][0]);
        }
    }
    return 0;
}



HDOJ 1914 The Stable Marriage Problem


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