PAT-Advanced-1004 Counting Leaves

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input

Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.
Output

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output “0 1” in a line.

Sample Input
2 1
01 1 02
Sample Output
0 1

解题思路

这题一开始还真没明白意思,以为算点的叶子数,后来才知道原来是算层的,一代一代的输出,知道要干什么就很简单了,直接用DFS搜索就可以,而且还不需要设置访问数组,因为是树不可能会有回路,要求很简单,只要判断有没有叶结点就行,如果有叶结点,这层的叶结点数增加。

 

 

 

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