一家人

题目描述

最近zzq交了一个新朋友叫cj，他们在聊天的时候发现500年前他们竟然是一家人！现在zzq想知道cj是他的长辈，晚辈，还是兄弟。

输入

输入包含多组测试数据。每组首先输入一个整数N（N<=10），接下来N行，每行输入两个整数a和b，表示a的父亲是b（1<=a,b<=20）。zzq的编号为1，cj的编号为2。
输入数据保证每个人只有一个父亲。

输出

对于每组输入，如果cj是zzq的晚辈，则输出“You are my younger”，如果cj是zzq的长辈，则输出“You are my elder”，如果是同辈则输出“You are my brother”。

样例输入

51 32 43 54 65 661 32 43 54 65 76 7

样例输出

You are my elderYou are my brother

题意概括：

         现在有两个人，zzq和cj，给你n行数据，每行输入两个整数a和b，表示a的父亲是b。zzq的编号为1，cj的编号为2。根据这些数据推出zzq和cj的辈分关系。
解题数量：

         从a=1和a=2处开始找出所给数据中所有关于二者的人物，辈分从小到大排列，存入两个数组中，比较这两个数组，找出数组中第一个相同的人物，代表二人共同的祖先，然后看祖先在那个人代表的数组里排的靠前，那个人的辈分就大。

代码：

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

int main()

{

         inta[1000],b[1000],c[1000],d[1000],i,j,k,k1,k2,e,l1,l2,m,n,t1,t,f1,f2;

        

         while(scanf("%d",&n)!=EOF)

         {

                   memset(c,0,sizeof(c));

                   memset(d,0,sizeof(d));

                   for(i=0;i<n;i++)

                   {

                            scanf("%d%d",&a[i],&b[i]);

                   }

                   k2=0;

                   for(i=0;i<n;i++)

                   {

                            if(a[i]==1&&b[i]==2)

                            {

                                     printf("Youare my elder\n");

                                     k2=1;

                            }

                            if(a[i]==2&&b[i]==1)

                            {

                                     printf("Youare my younger\n");

                                     k2=1;

                            }

                   }

                   if(k2==0)

                   {

                  

                            for(i=0;i<n;i++)

                            {

                                     //printf("%d%d\n",a[i],b[i]);

                            }

                            l1=1;l2=2;k=0;e=0;

                            for(i=0;i<n;i++)

                            {

                                     for(j=0;j<n;j++)//printf("a[i]==%d  %d\n",a[i],l1);

                                     {

                                    

                                               if(a[j]==l1)//关于1的人物存入数组c中

                                               {

                                                        l1=b[j];

                                                        c[k]=a[j];

                                                        k++;

                                                        c[k]=b[j];

                                                        k++;

                                                       

                                               }

                                               //printf("a[i]==%d  %d\n",a[i],l2);

                                               if(a[j]==l2)//关于2的人物存入数组d中

                                               {

                                                        l2=b[j];

                                                        d[e]=a[j];

                                                        e++;

                                                        d[e]=b[j];

                                                        e++;

                                                       

                                               }

                                     }

                            }

                            for(i=0;i<k;i++)

                            {

                            //      printf("%d ",c[i]);

                            }

                            //printf("\n\n");

                            for(i=0;i<e;i++)

                            {

                            //      printf("%d ",d[i]);

                            }

                            //printf("\n\n");

                            k1=0;

                            f1=0;

                            f2=0;

                            for(i=0;i<k;i++)//找出共同的祖先和他所在的位置

                            {

                                     for(j=0;j<e;j++)

                                     {

                                               if(c[i]==d[j])

                                               {

                                                        f1=i;

                                                        f2=j;

                                                        k1=1;

                                                        break;

                                               }

                                     }

                                     if(k1==1)

                                     break;

                            }

                            //printf("f1f2%d%d\n",f1,f2);

                            if(f1>f2)

                            {

                                     printf("Youare my elder\n");

                            }

                            if(f1==f2)

                            {

                                     printf("Youare my brother\n");

                            }

                            if(f1<f2)

                            {

                                     printf("Youare my younger\n");

                            }

                   }

         }

        

         return0;

}