nyoj 282 You are my brother

You are my brother

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

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

#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)//判断是否是父节点 
		{
			if(end==1)//如果是1的父节点  自加1 
			sum1++;
			else  //是2的父节点 
			sum2++;
			dfs(i,end);//继续搜索 
			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");
	else
	printf("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(1,1);
		dfs(2,2);
		shu();
	}
	return 0;
}