HDU Marriage is Stable (稳定婚姻匹配)

题目

Albert, Brad, Chuck are happy bachelors who are in love with Laura, Marcy, Nancy. They all have three choices. But in fact, they do have some preference in mind. Say Albert, he likes Laura best, but that doesn't necesarily mean Laura likes him. Laura likes Chuck more than Albert. So if Albert can't marry Laura, he thinks Nancy a sensible choice. For Albert, he orders the girls Laura > Nancy > Marcy.

For the boys:

Albert: Laura > Nancy > Marcy
Brad: Marcy > Nancy > Laura
Chuck: Laura > Marcy > Nancy

For the girls:

Laura: Chuck > Albert > Brad
Marcy: Albert > Chuck > Brad
Nancy: Brad > Albert > Chuck

But if they were matched randomly, such as

Albert <-> Laura
Brad <-> Marcy
Chuck <-> Nancy

they would soon discover it's not a nice solution. For Laura, she likes Chuck instead of Albert. And what's more, Chuck likes Laura better than Nancy. So Laura and Chuck are likely to come together, leaving poor Albert and Nancy.

Now it's your turn to find a stable marriage. A stable marriage means for any boy G and girl M, with their choice m[G] and m[M], it will not happen that rank(G, M) < rank(G, m[G]）and rank(M, G) < rank(M, m[M]).

Input
Each case starts with an integer n (1 <= n <= 500), the number of matches to make. The following n lines contain n + 1 names each, the first being name of the boy, and rest being the rank of the girls. The following n lines are the same information for the girls. Process to the end of file.

Output
If there is a stable marriage, print n lines with two names on each line. You can choose any one if there are multiple solution. Print "Impossible" otherwise. Print a blank line after each test.

Sample Input
3
Albert Laura Nancy Marcy
Brad Marcy Nancy Laura
Chuck Laura Marcy Nancy
Laura Chuck Albert Brad
Marcy Albert Chuck Brad
Nancy Brad Albert Chuck

Sample Output
Albert Nancy
Brad Marcy
Chuck Laura

思路

稳定婚姻，用map转换一下字符串就方便操作了；

代码实现

#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 0x3f3f3f3f
#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; }
map <string ,int > mp_mm,mp_gg;
const int maxn=505;
int gg[maxn][maxn],mm[maxn][maxn];
int gg_match[maxn],mm_match[maxn];
string ss[maxn],str;
int p[maxn];
int n;


void input () {
    mp_mm.clear ();
    mp_gg.clear ();
    int cnt=0;
    rep (i,1,n) {
        cin>>ss[i];
        mp_gg[ss[i]]=i;
        rep (j,1,n) {
           cin>>str;
           if (!mp_mm[str])  {
               mp_mm[str]=++cnt;
               ss[cnt+n]=str;
           }
           gg[i][j]=mp_mm[str];
        }
    }

    rep (i,1,n) {
        cin>>str;
        int a=mp_mm[str];
        rep (j,1,n) {
            cin>>str;
            int b=mp_gg[str];
            mm[a][b]=j;
        }
    }

}

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

    }
}

int main () {
    while (cin>>n) {
       input ();
       solve ();
       rep (i,1,n) {
           cout<<ss[i]<<" "<<ss[gg_match[i]+n]<<endl;
       }
       cout<<endl;
    }
    return 0;
}