【PAT甲】1004 Counting Leaves (30分)（多叉树存储遍历，统计每层叶节点数）

problem

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

• 一个家族中假设有n名成员，其中m个人有孩子，分别给出这m个人孩子的名单
• 求每一辈有几个成员是没有儿子的
• 即给出一棵树（共n点，m个非叶节点及其子节点），问每一层各有多少个叶子结点

solution

• 多叉树的存储和遍历：一是链表指针结构强行乱搞，二是邻接表记录每个节点的子节点，vectorG[110]或者Tree[110][10]都可以。
• 用book[i]表示每一层的叶子节点个数，dfs遍历一遍树并记录当前depth，每访问到一个叶子节点的时候都更新book[depth]++；

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>Tree[110];
int book[110], maxdep;
void dfs(int node, int dep){
	if(Tree[node].size()==0){
		book[dep]++;
		maxdep = max(maxdep,dep);
		return ;
	}
	for(int i = 0; i < Tree[node].size(); i++)
		dfs(Tree[node][i], dep+1);
}
int main(){
	int n, m;
	cin>>n>>m;
	for(int i = 0; i < m; i++){
		int node, k;
		cin>>node>>k;
		for(int j = 0; j < k; j++){
			int c;  cin>>c;
			Tree[node].push_back(c);
		}
	}
	dfs(1,0);//从根节点开始遍历，当前深度为0
	cout<<book[0];
	for(int i = 1; i <= maxdep; i++)
		cout<<" "<<book[i];
	return 0;
}


  

 

  
 
 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/108410691