问题 : You are my brother_you are my brother csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/liu344439833/article/details/80140973

本文介绍了一个简单的算法，用于解决两个个体之间的家族关系问题，通过比较他们追溯到共同祖先的路径来确定彼此是兄长、弟弟还是兄弟。

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问题 : You are my brother

时间限制: 1 Sec 内存限制: 128 MB

http://218.198.32.182/problem.php?cid=1067&pid=1

题目描述

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

样例输出

You are my elder
You are my brother

解题思路：

就是一个找父亲问题，然后比较父亲的大小。

源代码：

# include <stdio.h>
# include <string.h>

int main(void)
{
	int n, x, y;
	int a[2001], i;
	while (~ scanf("%d", &n))
	{
		memset(a, 0, sizeof(a));
		for (i = 0; i < n; i ++)
		{
			scanf("%d %d", &x, &y);
			a[x] = y;
		}
		int c = 1, sum = 0;
		while(a[c] != 0)
		{
			sum ++;
			c = a[c];
		}
		int sum1 = 0;
		c = 2;
		while(a[c] != 0)
		{
			sum1 ++;
			c = a[c];
		}
		if(sum > sum1)  
            printf("You are my elder\n");  
        else if(sum < sum1)  
            printf("You are my younger\n");  
        else  
            printf("You are my brother\n"); 
	}
	return 0;
}