HDU1522 稳定婚姻匹配 模板

Marriage is Stable

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1047    Accepted Submission(s): 563
Special Judge

Problem Description

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

 

Author

CHENG, Long

 

Source

ZOJ

题意：
n位男士，n位女士，没人对各个异性有一个排序，代表对他们的喜欢程度。任务是将男生和女生一一配对，使得男生v和女生u不存在以下情况：（1）男生v和女生u不是舞伴；（2）他们喜欢对方的程度都大于喜欢各自当前舞伴的程度。

输出最稳定配对，好像没有Impossible

代码：

//稳定婚姻匹配  详细：白书352页
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
using namespace std;
const int maxn=505;
int n;
int pref[maxn][maxn],fum[maxn],fuw[maxn],ord[maxn][maxn],nex[maxn];
//pref:编号为i的man第j喜欢的woman的编号;fum:编号为i的woman的未婚夫；
//ord:编号为i的woman第j喜欢的man;
map<string,int>vm,vw;
map<int,string>um,uw;
queue<int>q;//未订婚的男士队列
//订婚函数
void engage(int m,int w)
{
    int pm=fum[w];
    if(pm){      //woman现有未婚夫
        fuw[pm]=0;//抛弃pm
        q.push(pm);//pm加入未定婚男士队列
    }
    fuw[m]=w;
    fum[w]=m;
}
int main()
{
    char ch[102];
    while(~scanf("%d",&n)){
        vm.clear();vw.clear();um.clear();uw.clear();
        memset(pref,0,sizeof(pref));
        int cnt=0;
        for(int i=1;i<=n;i++){
            scanf("%s",ch);
            vm[ch]=i;um[i]=ch;
            for(int j=1;j<=n;j++){
                scanf("%s",ch);
                if(!vw[ch]){
                    vw[ch]=++cnt;
                    uw[cnt]=ch;
                }
                pref[i][j]=vw[ch];//编号为i的男士第j喜欢的人
            }
            nex[i]=1;//接下来应向排名为1的女士求婚
            fuw[i]=0;//没有未婚妻
            q.push(i);
        }
        for(int i=1;i<=n;i++){
            scanf("%s",ch);
            int k=vw[ch];
            for(int j=1;j<=n;j++){
                scanf("%s",ch);
                ord[k][vm[ch]]=j;//在编号为k的女士心中编号为vm[ch]的男士的排名
            }
            fum[k]=0;//没有未婚夫
        }
        while(!q.empty()){
            int m=q.front();
            q.pop();
            int w=pref[m][nex[m]++];//下一个求婚对象
            if(w==0) continue;
            if(!fum[w])//女士没有未婚夫，直接订婚
                engage(m,w);
            else if(ord[w][m]<ord[w][fum[w]])//代替女士的现任未婚夫
                engage(m,w);
            else q.push(m);//直接被拒绝，下次再来
            }
        while(!q.empty()) q.pop();
        for(int i=1;i<=n;i++){
            cout<<um[i]<<" "<<uw[fuw[i]]<<endl;
        }
    }
    return 0;
}