学步园

Little A gets to know a new friend, Little B, recently. One day, they realize that they are family 500 years ago. Now, Little A wants to know whether Little B is his
elder, younger or brother.

There are multiple test cases.

For each test case, the first line has a single integer, n (n<=1000). The next n lines have two integers a and b (1<=a,b<=2000) each, indicating b is the father of
a. One person has exactly one father, of course. Little A is numbered 1 and Little B is numbered 2.

Proceed to the end of file.

For each test case, if Little B is Little A’s younger, print “You are my younger”. Otherwise, if Little B is Little A’s elder, print “You are my elder”. Otherwise,
print “You are my brother”. The output for each test case occupied exactly one line.

5
1 3
2 4
3 5
4 6
5 6
6
1 3
2 4
3 5
4 6
5 7
6 7

You are my elder
You are my brother

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int pre[2007],n;
int find(int x)
{
    int t=1;
    while(x!=pre[x])
    {
        x=pre[x];
        t++;
    }
    return t;
}
int main()
{
    freopen("D:\\test.txt","r",stdin);
    while(cin>>n)
    {
        int a,b,i;
        memset(pre,0,sizeof(pre));
        for(i=1;i<=n;i++)
        {
            cin>>a>>b;
            pre[a]=b;
        }
        int x=find(1);
        int y=find(2);
        if(x<y)
        cout<<"You are my younger"<<endl;
        else if(x>y)
        cout<<"You are my elder"<<endl;
        else cout<<"You are my brother"<<endl;
    }
    return 0;
}