稳定婚姻问题

It is commonly stated as: Given n men and n women, where each person has ranked all members of the opposite sex with a unique number between 1 and n in order of preference, marry the men and women off such that there are no two people of opposite sex who would both rather have each other than their current partners. If there are no such people, all the marriages are "stable".

http://hi.baidu.com/fandywang%5Fjlu/blog/item/7c20b0d4c9149207a18bb706.html

Algorithm function stableMatching {    Initialize all m ∈ M and w ∈ W to free    while ? free man m who still has a woman w to propose to {       w = m's highest ranked such woman       if w is free         (m, w) become engaged       else some pair (m', w) already exists         if w prefers m to m'           (m, w) become engaged           m' becomes free         else           (m', w) remain engaged    }}以上算法是 male-optimal We say that the marriage between man A and woman B is feasible if there exists a stable pairing in which A and B are married. When we say a pairing is male-optimal, we mean that every man is paired with his highest ranked feasible partner. Similarly, a female-pessimal pairing is one in which each female is paired with her lowest ranked feasible partner. Ladies' Choice#include <iostream>
#include <queue>
using namespace std;
short girlboylikegirl[1001][1001],boyparterboylikegirl[1001][1001],girlparter[1001],ba[1001];
short boylikegirl[1001][1001],cur[1001];
bool boyfree[1001];
queue<int> q;
int main()
{
int p,n,i,j;
scanf("%d",&p);
while(p--)
{
   scanf("%d",&n);
   for(i=1;i<=n;i++)
   {
    for(j=0;j<n;j++)
     scanf("%d",&girlboylikegirl[i][j]);
   }
   for(i=1;i<=n;i++)
   {
    for(j=0;j<n;j++){
     scanf("%d",&boyparterboylikegirl[i][j]);
     boylikegirl[i][boyparterboylikegirl[i][j]]=j;
    }
   }
   memset(boyfree,0,sizeof(boyfree));
   memset(cur,0,sizeof(cur));
   for(i=1;i<=n;i++)
    q.push(i);
   while(!q.empty())
   {
    i=q.front();
    q.pop();
    for(j=cur[i];j<n;j++)
    {
     if(boyfree[girlboylikegirl[i][j]]==0)
     {
      girlparter[i]=girlboylikegirl[i][j];
      ba[girlboylikegirl[i][j]]=i;
      boyfree[girlboylikegirl[i][j]]=1;
      cur[i]=j+1;
      break;
     }
     else if(boylikegirl[girlboylikegirl[i][j]][i]<boylikegirl[girlboylikegirl[i][j]][ba[girlboylikegirl[i][j]]])
     {
      girlparter[i]=girlboylikegirl[i][j];
      q.push(ba[girlboylikegirl[i][j]]);
      ba[girlboylikegirl[i][j]]=i;
      cur[i]=j+1;
      break;
     }
    }
   }
   for(i=1;i<=n;i++)
    printf("%d\n",girlparter[i]);
}
return 0;
}