[1218] You are my brother

[1218] You are my brother

• 时间限制: 1000 ms　内存限制: 131072 K
• 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

  题目大意：输入N对数a，b，表示b是a的父亲，最后求的是1和2的关系，也就是看1和2与根的关系（注意，若a和b不在同一棵树上，则他们的关系为brother）

• 思路：题目中说不存在两个父亲，所以我们可以定义一个数组，把一个数存到这个数上，形如a[1]=3,a[3]=5……，这样每个数就都会连到一起，最后循环找一下根深度就好了，比赛的时候数组开大了，一直超时…………

代码：
 

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f3
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=2e7+100;
const double pi=acos(-1.0);
const int N=2e5+10;
const int mod=1e9+7;
int a[N];
int main()
{

    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(int i=0; i<n; i++)
        {

            int aa,bb;
            scanf("%d%d",&aa,&bb);
            a[aa]=bb;
        }
        int ans1=0,ans2=0;
        int f1=1,f2=2;
        while(a[f1])
        {
            ans1++;
            f1=a[f1];
        }
        while(a[f2])
        {
            ans2++;
            f2=a[f2];
        }
        if(ans1==ans2)
            printf("You are my brother\n");
        else if(ans1>ans2)
            printf("You are my elder\n");
        else
            printf("You are my younger\n");
    }
    return 0;
}