## You are my brother

​ 时间限制：1000 ms | 内存限制：65535 KB

​ 难度：3

### 描述

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.

### 输入

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.

### 输出

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.

### 样例输入

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

### 样例输出

You are my elder
You are my brother

### 代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<stack>
#include<algorithm>
#define MAX 2001
using namespace std;
int set[MAX];
int maxnode;
int sum1,sum2;
int max(int x,int y)
{
return x>y?x:y;
}
void dfs(int p,int pos)
{
int i;
if(set[p]==p)//找到根节点
return ;
for(i=1;i<=maxnode;i++)
{
if(set[p]==i)//找到父节点
{
if(pos==1)
sum1++;
else
sum2++;
dfs(i,pos);//继续
}
}
}
void init()
{
int i;
for(i=1;i<MAX;i++)
set[i]=i;
}
void solve()
{
if(sum1==sum2)
printf("You are my brother\n");
else if(sum1>sum2)
printf("You are my elder\n");
else
printf("You are my younger\n");
}
int main()
{
int n,i,j,x,y;
while(scanf("%d",&n)!=EOF)
{
init();
maxnode=0;
while(n--)
{
scanf("%d%d",&x,&y);
maxnode=max(max(x,y),maxnode);
set[x]=y;
}
sum1=sum2=0;
dfs(1,1);
dfs(2,2);
solve();
}
return 0;
}