2019-02-25-The Stable Marriage Problem-POJ3487-稳定婚姻算法

POJ-3487-The Stable Marriage Problem

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.

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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.

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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

题目简述

对于给定的n对男女，再给出相应的好感度排名，即男生眼中女生的好感度排名(n行)和女生眼中男生的好感度排名(n行)。

对于两个匹配a-b，c-d，如果a的眼中d的排名比b高，d的眼中a排名比c高，那么这样的匹配不稳定。

试求出一种匹配方式，使得对于所有的男生和女生，都是一个稳定的匹配。

条目检索：稳定婚姻算法

输入输出

输入：无难点，字符串读入即可

输出：无难点

使用模板（Gale_Shapley稳定婚姻算法）

struct Gale_Shapley{
	static const int N=60;
	int n;
	int fg;
	int woman[N][N];        //第i个girl对编号j的boy的好感排位 
	int man[N][N];          //第i个boy第j喜欢的girl 
	int man_match[N],woman_match[N];    //已匹配的男女 
	int p[N];               //第i个boy已经表白的人数 
	void solve(){
		for(int i=0;i<N;i++)p[i]=1;
		memset(man_match,0,sizeof(man_match));
		memset(woman_match,0,sizeof(woman_match));
		fg=true;
		while(fg){
			fg=false;
			for(int i=1;i<=n;i++){
				if(!man_match[i]){
					int t=man[i][p[i]++];
					if(!woman_match[t]){
						man_match[i]=t;
						woman_match[t]=i;
					}else if(woman[t][woman_match[t]]>woman[t][i]){
						man_match[woman_match[t]]=0;
						man_match[i]=t;
						woman_match[t]=i;
					}
					fg=true;
				}
			}
		}
	}
}T;

AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
//使用模板
struct Gale_Shapley{
	static const int N=60;
	int n;
	int fg;
	int woman[N][N];//第i个girl对编号j的boy的好感排位 
	int man[N][N];//第i个boy第j喜欢的girl 
	int man_match[N],woman_match[N];//已匹配的男女 
	int p[N];//第i个boy已经表白的人数 
	void solve(){
		for(int i=0;i<N;i++)p[i]=1;
		memset(man_match,0,sizeof(man_match));
		memset(woman_match,0,sizeof(woman_match));
		fg=true;
		while(fg){
			fg=false;
			for(int i=1;i<=n;i++){
				if(!man_match[i]){
					int t=man[i][p[i]++];
					if(!woman_match[t]){
						man_match[i]=t;
						woman_match[t]=i;
					}else if(woman[t][woman_match[t]]>woman[t][i]){
						man_match[woman_match[t]]=0;
						man_match[i]=t;
						woman_match[t]=i;
					}
					fg=true;
				}
			}
		}
	}
}T;

int main(){
	int cases;
	scanf("%d",&cases);
	while(cases--){
		scanf("%d",&T.n);
		char boy[27];         //编号对应的字母
		char girl[27];        //编号对应的字母
		char s[30];
		map<char,int>bo,gi;   //字母对应的编号
        //建立映射
		for(int i=1;i<=T.n;i++){
			scanf("%s",s);
			boy[i]=s[0];
			bo[s[0]]=i;
		}
		for(int i=1;i<=T.n;i++){
			scanf("%s",s);
			girl[i]=s[0];
			gi[s[0]]=i;
		}
        //读入排名
		for(int i=1;i<=T.n;i++){
			scanf("%s",s);
			int x,y;
			x=bo[s[0]];
			for(int j=1;j<=T.n;j++){
				y=gi[s[1+j]];
				T.man[x][j]=y;      //男生x所第j喜欢的女生是y
			}
		}
		for(int i=1;i<=T.n;i++){
			scanf("%s",s);
			int x,y;
			x=gi[s[0]];
			for(int j=1;j<=T.n;j++){
				y=bo[s[1+j]];
				T.woman[x][y]=j;    //对于女生x而言男生y的排名是j
			}
		}

		T.solve();

		for(int i=1;i<=T.n;i++){
			printf("%c %c\n",boy[i],girl[T.man_match[i]]);
		}
		if(cases!=0)printf("\n");
	}
	return 0;
}

参考链接

https://blog.youkuaiyun.com/u011815404/article/details/81395124

https://blog.youkuaiyun.com/cscmaker/article/details/8291131