poj3487 The Stable Marriage Problem（最稳定的婚姻）
The Stable Marriage Problem
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 1966 Accepted: 838
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:abcSample Output
a A
b B
c C
a B
b A
c C有n个男士n个女士，每个人都对所有异性有一个评价值，男的对女的表白，如果女的没有对象，就会接纳这个男的，如果女的有对象但是女的对这个对象的好感不如表白这个人的好感，这个女的会抛弃现在的对象和对她表白的那个人在一起，要求出最稳定的婚姻状态。
[cpp]
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std; int male[30],female[30];
int m[30][30],fm[30][30]; int vis[30]; int n;
char t[60]; queue<int> q; void scanf_name()
{ char tmp[30]; gets(t);
//cout<<t <<endl; for(int i=0; i<n; i++)
{ gets(tmp);
//cout<<tmp <<endl;
int poit=tmp[0]-'a'+1;
q.push(poit);
for(int j=2; j<strlen(tmp); j++)
{ int k=tmp[j]-'A'+1;
m[poit][j-1]=k; } }
for(int i=0; i<n; i++) { //cout<<"cin"<<endl;
gets(tmp); int poit=tmp[0]-'A'+1; for(int j=2; j<strlen(tmp); j++)
{
int k=tmp[j]-'a'+1;
fm[poit][k]=j-1;
} } } void GaleShapley()
{ while(!q.empty())
{ int poit=q.front();
//第poit位男士
q.pop();
//if(male[poit]!=-1)
//continue;
if(vis[poit]>n)
continue;
int i=++vis[poit];
//第poit位男士最喜欢的第i位女士
//vis[poit]++; int fem=m[poit][i];
//第poit位男士最喜欢的第i为女士是第fem位
if(female[fem]==-1)
//如果第fem位女士没有喜欢的人 {
female[fem]=poit;
male[poit]=fem;
continue; }
else {
int temp=female[fem];
if(fm[fem][temp]>fm[fem][poit])
//如果第fem位女士对当前喜欢的人的喜欢不如对第poit位男士的喜欢 {
female[fem]=poit;
male[poit]=fem;
male[temp]=-1;
q.push(temp);
continue;
} else
{ q.push(poit);
continue;
} }
} } int main() {
int s; scanf("%d",&s);
while(s--) {
memset(vis,0,sizeof(vis));
memset(male,-1,sizeof(male));
memset(female,-1,sizeof(female));
while(!q.empty()) q.pop();
scanf("%d&qu
补充：软件开发 , C++ ,
