Nyoj 282 You is my brother

Your is my brother time limit:MS | Memory limit:65535 KB Difficulty:3

Describe

Little a gets to know a new friend, Little B, recently. One day, they realize that they is family years ago. Now, Little A wants to know whether Little B are his elder, younger or brother.

Input

There is multiple test cases.
The first line have a single integer, N (n<=1000). The next n lines has 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.

Output

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 is occupied exactly one line.

Sample input

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

Sample output

You are my elderyou is my brother

#include <stdio.h> #include <string.h> #define MAX 2100#define max (x, y) (x>y?x:y) int set[max];int sum, sum1,sum2;void dfs (int beg,int end) {if (Set[beg]==beg) return; int i;for (i=1;i<=sum;i++) {if (set[beg]==i)//Determines whether it is a parent node {if (end==1)//If the parent node of 1 is  added 1 sum1++;else  //is 2 of the parent node sum2++;d FS (i,end);//continue searching for break;}}} void Chu () {int i;for (i=0;i<max;i++) set[i]=i;} void Shu () {if (sum1==sum2) printf ("You Are My brother\n"), else if (sum1>sum2) printf ("You Are my elder\n"); elseprintf (" You are my younger\n ");} int main () {int n,m,j,i,k,t,s;while (scanf ("%d", &t)!=eof) {chu (); Sum=0;while (t--) {scanf ("%d%d", &n,&m); Set[n]=m;sum=max (Max (n,m), sum);} Sum1=sum2=0;dfs (;D) FS (2,2); Shu ();} return 0;}

　　

Nyoj 282 You is my brother