【笨方法学PAT】1004 Counting Leaves（30 分）_pat counting leaves-优快云博客

本文介绍了一种算法，用于解决一个特定的问题：给定一个家族树结构，通过深度优先搜索（DFS）算法来计算每层的叶子节点数量。该问题通过构建树的邻接矩阵表示，并采用递归方式遍历树结构来实现。

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一、题目

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

二、题目大意

给一个树，求每一层叶子节点的个数

三、考点

树、DFS

四、解题思路

1、使用邻接矩阵保存树的结构，将每个儿子节点保存到父节点的 vector 中，使用 int 数组 level_num 保存每一层叶节点的个数；

2、DFS搜索， dfs(int index,int level) 维护当前节点和层数两个变量；

3、DFS中遍历下一层的时候，需要注意当前节点的数字不是 i ，而是 vec[index][i]。这里是使用邻接矩阵，如果直接使用二维数组的话，思路简单些，但占用内存会增加；

4、在输出的过程中，level_num为 0 的情况下也要输出，而最大层数也不知道，所以要有一个全局变量 max_level 维护最大层数。

【注意】数据保存在二维数组、邻接矩阵、链表中遍历的方式有些不同，不要混淆。

五、代码

#include<iostream>
#include<vector>
#define N 110
using namespace std;

vector<int> vec[N];
int n, m;
int level_num[N] = {0};
bool visit[N] = { false };
int max_level = 0;

void dfs(int index,int level) {
	//更新最大值
	if (max_level < level)
		max_level = level;

	//终止条件
	if (vec[index].size() == 0) {
		level_num[level]++;
	}

	//下一层
	for (int i = 0; i < vec[index].size(); ++i) {
		int now_index = vec[index][i];
		if (visit[now_index] == false) {
			visit[now_index] = true;
			dfs(now_index, level + 1);
		}
	}
}

int main() {
	//输入
	cin >> n >> m;
	while (m--) {
		int id, k,a;
		cin >> id >> k;
		while (k--) {
			cin >> a;
			vec[id].push_back(a);
		}
	}

	//n==0特殊情况处理
	if (n == 0)
		return 0;

	//DFS
	visit[1] = true;
	dfs(1,0);

	//输出
	for (int i = 0; i <= max_level; ++i) {
		if (i != 0)
			cout << " ";
		cout << level_num[i];
	}

	system("pause");
	return 0;
}