1004. Counting Leaves (30)-优快云博客

本文介绍了一种算法，用于解决计数家族树中各层级的叶节点数量的问题。通过构建树的数据结构并采用层序遍历的方法，实现对输入的家族树进行解析，并输出每个层级的叶节点数目。

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1004. Counting Leaves (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

问题分析：题目的内核是数每层的叶节点，算法核心是树的层序遍历，利用类似于六度空间的tail，last两个变

量标记一层结束，树的表示方式为first child，Next Sibling，存储方式是将所有节点放在结构数组中，方便利用下标访问对应编号的节点。#include<stdio.h>
#include<stdlib.h>
typedef struct NODE *PtrN;
struct NODE
{
	int id;
	int fc;//first child
	int ns;//next sibling
};
typedef struct QUE *PtrQ;
struct QUE
{
	int front;
	int rear;
	int* que;
	int size;
};
PtrQ CreateQue(int N)
{
	PtrQ Que = (PtrQ)malloc(sizeof(struct QUE));
	Que->size = N;
	Que->front = 0;
	Que->rear = 0;
	Que->que = (int*)malloc(sizeof(int)*N);
	return Que;
}
void AddQue(PtrQ Que, int data)
{
	if ((Que->rear + 1) % Que->size == Que->front)return;
	Que->rear = (Que->rear + 1) % Que->size;
	Que->que[Que->rear] = data;
	return;
}
int Isempty(PtrQ Que)
{
	if (Que->front == Que->rear)return 1;
	else return 0;
}
int DeQue(PtrQ Que)
{
	int ans;
	if (Isempty(Que))return -1;
	Que->front = (Que->front + 1) % Que->size;
	return Que->que[Que->front];
}
int main()
{
	
	
	int N, M;
	int i, j, k,m,temp,tmp;
	
	scanf( "%d %d", &N, &M);
	PtrN List = (PtrN)malloc(sizeof(struct NODE)*N);
	for (i = 0; i < N; i++)
	{
		List[i].id = i + 1;
		List[i].fc = -1;
		List[i].ns = -1;
	}
    
	for (i = 0; i < M; i++)
	{
		scanf("%02d %d", &k,&temp);
		for (j = 0; j < temp; j++)
		{
			scanf( "%02d", &tmp);
			if (j == 0)
			{
				List[k - 1].fc = tmp - 1;
				m = tmp - 1;
			}
			else
			{
				List[m].ns = tmp - 1;
				m = tmp - 1;
			}
		}

	}
	int cnt = 0;
	int last = 0;
	int tail = 0;
	PtrQ Que;
	Que = CreateQue(N);
	if (List[0].fc == -1)
	{
		printf("1\n");
	}
	else AddQue(Que, 0);

	while (!Isempty(Que))
	{
		temp = DeQue(Que);
		i = List[temp].fc;
		if (i == -1)cnt++;
		else
		{
			while(i!=-1)
			{
				AddQue(Que, i);
				tail = i;
				i = List[i].ns;
				
			}
		}
		if (temp == last)
		{
			if (temp == 0)
				printf("%d", cnt);
			else
				printf(" %d", cnt);
			last = tail;
			cnt = 0;
		}
		
	}

	
	return 0;
}