poj3487(Gale-Shapley算法) The Stable Marriage Problem(稳定婚姻问题)

2017年05月03日 13点热度 0人点赞 0条评论
The Stable Marriage Problem
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3011   Accepted: 1277

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

男士是小写,女士是大写

#include
#include
#include
#include
#include
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 freemale;//没有配对的男士 
		scanf("%d",&couple);
		for(int i=0;ifemalelike[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,否则死讯坏或者错误 
				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

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update

纸上得来终觉浅, 绝知此事须躬行。