PAT-A1004 Counting Leaves 题目内容及题解

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

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0

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.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

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

题目大意

题目给出一个家谱树，格式为ID 孩子数 孩子ID，从根节点所在层开始输出每层中没有孩子节点的节点（叶子节点）的数量。

解题思路

1. 建立结构体数组存储每个节点；
2. 初始化题目所给的树；
3. 采用DFS的方法将所有节点所在层记入结点结构体中（以根节点所在层为第0层），并记录最大层数；
4. 遍历节点，找出每层的叶子节点并计数；
5. 输出结果并返回0值。

代码

#include

#define MAXN 110
struct Node{
    int height;
    int num;
    int children[MAXN];
}node[MAXN];

int N,M;
int H[MAXN],max=-1; 
 
void DFS(int root,int height){
    int i;
    node[root].height=height;
    for(i=0;i

运行结果