PAT甲级1004(C语言)

1004 Counting Leaves （30 分)

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<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.

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

 

来自

题目思路:

    题目意思为求得一颗数的每层叶节点个数，根据题目给出的输入数据。想到可以采用变长动态数组来存储数据。

这里采用数组单链表来存储数据(c++中vector类模板也可实现相同功能)。

数据结构形式为：

#define max 100

typedef struct node{

Int data;

node* next;

}node;

node p[max];

dfs:每向下递归一层，层次等级+1，并且更新根结点，返回递归时，返回原状态，并且在原状态上指针后移到下一位。

代码如下：

#include
#include
#include
#define max 100
typedef struct node{
	int data;
	struct node *next;
}node;
node p[max];//下标表示根节点 

int num[max];//各层次叶节点数 
int maxLevel = -1;//最大深度 
int main() {
    void dfs(int level, int root);
	int n, m;//树中节点个数，非叶节点个数
	/*初始化数组链表*/ 
	for (int i = 0; i < max; ++i) {
	p[i].data = 0;
	p[i].next = NULL;
	}
	scanf("%d %d", &n, &m);
	memset(num, 0, max);//初始化各层次节点数 
	for (int i = 0; i < m; ++i) {
		int root, k;
		scanf("%d %d", &root, &k);
		node *temp = &p[root];
		/*构建数组链表*/ 
		for (int j = 0; j < k; ++j) {
			node *q = (node*)malloc(sizeof(node));
			scanf("%d", &q->data);
			
			q->next = temp->next;
			temp->next = q;
			temp = temp->next;
		}
		p[root].data = k;//存储孩子节点个数 
		temp->next = NULL;
	}
	dfs(0, 1);
	
	printf("%d", num[0]);
	for (int i = 1; i <= maxLevel; ++i) {
		printf(" %d", num[i]);
	}
	return 0;

	
} 

void dfs(int level, int root)
{
	node *temp = &p[root];
	
	if (temp->data == 0) {
		num[level]++;
		if (level > maxLevel) maxLevel = level;
		return ;
	}
	
	for (int i = 0; i < temp->data && temp->next != NULL; ++i) {
		dfs(level+1, temp->next->data);
    	        temp = temp->next;
	}
}