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]).
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.
<b< dd="">
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.
<b< dd="">
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
<b< dd="">
Albert Nancy
Brad Marcy
Chuck Laura
题目大意：给出几个男孩喜欢几个女孩的程度，再给出女孩喜欢男孩的程度，现在需要对着几个男孩还有女孩进行配对，要求达到皮队合适。
解题思路：稳定婚姻问题，该问题按照思路大致是这样：以女生为主，进行匹配。首先，根据男生都向自己喜欢的女生进行求婚，每个女生男生在向自己求婚的男生中，选取一个自己最喜欢的，如果只有一个，那就选择这个男生。之后如果还有男生是单身的，再想自己次要喜欢的女生进行求婚，如果这个当前女生已经有了对象，并且当前女生对目前对象喜欢程度不如现在求婚的，那该女生可以换对象，她之前的对象就成了单身。就这样一次进行匹配，因为男女生人数相等，这样进行匹配的最终结果是每个男生都有一个女生，每个女生也都有一个男生。并且匹配最坏的情况下，男生也能选择一个最喜欢自己的女生。
代码：
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<map>
#include<algorithm>
using namespace std;
#include<queue>
int man[510][510];
int G[510][510];
int book[510][510];
int a[510];
int b[510];
string s1[510];
string s2[510];
char s0[110];
int main()
{
int i,j,k,x,y,z,v,n;
int d[510];
while(scanf("%d",&n)!=EOF)
{
map<string,int> s3,s4;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(man,0,sizeof(man));
memset(book,0,sizeof(book));
memset(G,0,sizeof(G));
scanf("%s",s0);
s3[s0]=1;
s1[1]=s0;
for(i=1;i<=n;i++)
{
scanf("%s",s0);
s4[s0]=i;
s2[i]=s0;
man[1][i]=i;
}
for(i=2;i<=n;i++)
{
scanf("%s",s0);
s3[s0]=i;
s1[i]=s0;
for(j=1;j<=n;j++)
{
scanf("%s",s0);
man[i][j]=s4[s0];
}
}
for(i=1;i<=n;i++)
{
scanf("%s",s0);
x=s4[s0];k=n;
for(j=1;j<=n;j++)
{
scanf("%s",s0);
G[x][s3[s0]]=k--;
}
}
queue<int> p;
for(i=1;i<=n;i++)
p.push(i);
while(!p.empty())
{
y=p.front();
p.pop();
for(i=1;i<=n;i++)
{
z=man[y][i];
if(!book[y][z]&&z)
break;
}
book[y][z]=1;
if(!b[z])
{
b[z]=y;
a[y]=z;
}
else
{
if(G[z][b[z]]<G[z][y])
{
p.push(b[z]);
b[z]=y;
a[y]=z;
}
else
p.push(y);
}
}
for(i=1;i<=n;i++)
cout<<s1[i]<<' '<<s2[a[i]]<<endl;
printf("\n");
}
return 0;
}错误分析：stl的map了解不清，运用错误，画了好久才找到错误原因。