Marriage is Stable

Marriage is Stable

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 64 Accepted Submission(s): 43

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

Recommend 8600

/*
题意：给你n个男生暗恋的对象，n个女生暗恋的对象，如果刚好能组成n对不重复的情侣，就输出，如果不可能的话，就输出Impossible

初步思路：很典型的二分匹配问题
*/
#include<bits/stdc++.h>
using namespace std;
/***********************二分匹配模板**************************/
const int MAXN=1000;
int g[MAXN][MAXN];//编号是0~n-1的 
int linker[MAXN];//记录匹配点i的匹配点是谁
bool used[MAXN];
map<string,int> m;
map<int ,string> M;
int len=3;
int n;
string bname,gname;
bool dfs(int u)//回溯看能不能通过分手来进行匹配
{
    int v;
    for(v=0;v<n*2;v++)
        if(g[u][v]&&!used[v])
        //如果有这条边，并且这条边没有用过
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))//如果这个点没有匹配过，并且能找到匹配点，那么就可以以这个边作为匹配点
            {
                linker[v]=u;
                return true;
            }    
        }  
    return false;  
}    
int hungary()//返回最大匹配数
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<n*2;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))//如果这个点有匹配点 
            res++;
    } 
    return res;   
}
/***********************二分匹配模板**************************/
void init(){
    len=2;
    memset(g,0,sizeof g);
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=0;i<n;i++){
            cin>>bname;
            m[bname]=i;
            M[i]=bname;
            for(int j=0;j<n;j++){
                cin>>gname;
                if(m.find(gname)==m.end()){
                    m[gname]=++len;
                    M[len]=gname;
                }
                g[m[bname]][m[gname]]=1;
            }
        }
        // for(int i=0;i<n*2;i++){
            // cout<<M[i]<<" ";
        // }cout<<endl;
        
        for(int i=0;i<n;i++){
            cin>>gname;
            for(int j=0;j<n;j++){
                cin>>bname;
                g[m[gname]][m[bname]]=1;
            }
        }
        // for(int i=0;i<n*2;i++){
            // for(int j=0;j<n*2;j++){
                // cout<<g[i][j]<<" ";
            // }cout<<endl;
        // }
        //cout<<hungary()<<endl;
        if(hungary()==n*2){
            for(int i=0;i<n;i++){
                cout<<M[i]<<" "<<M[linker[i]]<<endl;
            }
        }else{
            puts("Impossible");
        }
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wuwangchuxin0924/p/6421682.html