**题目描述
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**
题意就是 输入n行，每行两个整数，表示前者是儿子，后者是父亲，第一行输入A的关系，第二行输入B的关系，最后他们的祖先必然相同，如果A的节点数大于B的，则说明You are my elder，相等时You are my brother，小于时You are my younger。
1：由于是单源路径，所以可以只用一个数组存储节点之间的关系。
2：数组初始化为自身是自身的父亲，下标为儿子，存储的元素为父亲。
3：在一个循环内，如果自身不是自身的父亲，既不是他们的共同祖先时，就继续往后查找，sum变量和count分别记录A和B的节点数，最后进行比较即可。
(为什么这。。。让我想起了并查集)
#include<stdio.h>
#include<string.h>
int f[2000];
int main()
{
int sum,count;
int i,n,a,b,k,l;
while(scanf("%d",&n)!=EOF)
{
memset(f,0,sizeof(f));
for( i = 1; i <= n; i ++)//步骤2
f[i] = i;
for( i =1; i <= n; i ++)
{
scanf("%d%d",&a,&b);
f[a] = b;
}
k = 1;
sum = 0;
while(f[k] != k)//步骤3
{
sum ++;
k = f[k];
}
l = 2;
count = 0;
while(f[l] != l)
{
count++;
l = f[l];
}
if( sum < count)
printf("You are my younger
");
else if( sum == count)
printf("You are my brother
");
else
printf("You are my elder
");
}
return 0;
}错误点：
学校的OJ让我很心碎啊 明明数组开到了1100就满足题意，它却显示运行错误，我改到2000就AC了
后记：这道题明明不初始化置0也没毛病啊，然而，它直接显示时间超限，今天练习赛的一道Kruskal纯模板题，我手动quicksort就导致超限，学长说直接用C++模板库里的sort就行，现在太菜，还分不清这些排序的时间复杂度的区别