PAT （A ）1004. Counting Leaves (30)_where id is a two-digit number representing a give-优快云博客

本文链接：https://blog.youkuaiyun.com/firogao/article/details/75116107

本文介绍了一种通过遍历家族树来统计每级无子女成员数量的算法。输入包括节点总数及非叶节点数，输出为各层级无子女成员的数量。文章提供了完整的C++实现代码，并附带了示例输入输出。

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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

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树结构 遍历

Errors：1.queue的front（）不叫pop （）

	2.vector 删除第一位元素方法

#include<cstdio> 
#include<vector>
#include <algorithm>
#include <iostream>
#include <cstring> 
#include<queue>
using namespace std;

int main()  
{  
  int nodes;
  int internal;
  int children[101][101];
  cin>>nodes>>internal;
  int n=internal;
  int num[101]={0};
  int now;
  int childrennum;
  vector<int>  ans;
  while(n!=0){
  	cin>>now>>childrennum;
  	num[now]=childrennum;
  	for(int i =1;i!=childrennum+1;i++){
  		cin>>children[now][i];	 
	  }
  	n--;
  }
  
 
  queue<int> next;
  int sum;
  next.push(1);
  while(next.size()!=0){
  	sum=0;
  	 queue<int> now;
	  while(next.size()!=0){
	  	int i=next.front();
	  
	  	if(num[i]==0) {
		sum++;
		  }
	  	else{
	  		for(int k=1;k!=num[i]+1;k++){
			  
	  			now.push(children[i][k]);
	  		
			  }
		  }
		  next.pop();
	
	  }
	  
	  	ans.push_back(sum);
		next=now;
  }

 for(int l=0;l!=ans.size();l++){
	 if(l!=ans.size()-1)	cout<<ans[l]<<" " ;
	 else	  cout<<ans[l]<<endl ;
	 	
  } 

  return 0;
}