问题描述

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

提示

无

来源

辽宁省赛2010

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<vector>

#include<math.h>

#include<cstdio>

#include<sstream>

#include<numeric>//STL数值算法头文件

#include<stdlib.h>

#include <ctype.h>

#include<string.h>

#include<iostream>

#include<algorithm>

#include<functional>//模板类头文件

using namespace std;

typedef long long ll;

const int maxn=11000;

const int INF=0x3f3f3f3f;

int a,b,n;

int fa[maxn];

int main()

{

    while(cin>>n)

    {

        int x1,x2,cot1,cot2;

        for(int i=0; i<maxn; i++)

            fa[i]=i;

        while(n--)

        {

            scanf("%d %d",&a,&b);

            fa[a]=b;

        }

        x1=1;

        cot1=0;

        while(fa[x1]!=x1)

        {

            x1=fa[x1];

            cot1++;

        }

        x2=2;

        cot2=0;

        while(fa[x2]!=x2)

        {

            x2=fa[x2];

            cot2++;

        }

        if(cot1>cot2)

            printf("You are my elder\n");

        else if(cot1<cot2)

            printf("You are my younger\n");

        else

            printf("You are my brother\n");

    }

    return 0;

}