題面

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

思路

裸題不說了，但我又忍不住吐槽HDU的評測機，在題庫交的PE程式碼在專題訓練裡面可以ac，專題訓練裡面ac的程式碼，題庫裡PE，呵呵~

程式碼實現

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=0;i<end;i++)
#define inf 0x3f3f3f3f3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f = -1;
        ch=getchar();
    } 
    while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }   return x*f;
}
const int maxn=30;
int t,n;
string s;
int cnt;
char a[maxn],b[maxn];
map <char,int > mp_mm,mp_gg;
int gg[maxn][maxn],mm[maxn][maxn];
int gg_match[maxn],mm_match[maxn];
int p[maxn];
bool sign=0;

void Gs () {
   bool flag=1;
   MT (gg_match,-1);
   MT (mm_match,-1);
   MT (p,0);
   while (flag) {
      flag=0;
      rep (i,1,n) {
          if (gg_match[i]==-1&&p[i]<=n) {
              int c=gg[i][++p[i]];
              if (mm_match[c]==-1) {
                  mm_match[c]=i;
                  gg_match[i]=c;
              }
              else if (mm[c][i]<mm[c][mm_match[c]]) {
                  gg_match[mm_match[c]]=-1;
                  gg_match[i]=c;
                  mm_match[c]=i;
              }
              flag=1;
          }
      }
   }
}

void countrank () {
   rep (i,1,n) {
       cin>>s;
       cnt=0;
       int now=mp_gg[s[0]];
       rep (j,2,s.size ()-1) {
           int girl=mp_mm[s[j]];
           gg[now][++cnt]=girl;
       }
   }
   rep (i,1,n) {
       cin>>s;
       cnt=0;
       int now=mp_mm[s[0]];
       rep (j,2,s.size ()-1) {
           int boy=mp_gg[s[j]];
           mm[now][boy]=++cnt;
       }
   }
}

void solve () {
   cin>>n;
   cnt=0;
   mp_mm.clear ();
   mp_gg.clear ();
   rep (i,1,n) {
       cin>>a[i];
       if (mp_gg[a[i]]==0) {
          mp_gg[a[i]]=(++cnt);
       }
   }
   cnt=0;
   rep (i,1,n) {
       cin>>b[i];
       if (mp_mm[b[i]]==0) {
           mp_mm[b[i]]=(++cnt);

       }
   }
   countrank ();
   Gs ();
//    rep (i,1,n) { 
//     rep (j,1,n) {
//         cout<<gg[i][j]<<" ";
//     }
//     cout<<endl;
//    }
   rep (i,1,n) {
       cout<<a[i]<<" "<<b[gg_match[i]]<<endl;
   }
}

int main () {
   cin>>t;
   int ca=0;
   while (t--) {
       solve ();
       printf ("\n");
   }
    return 0;
}