1004 Counting Leaves

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#PAT (Advanced Level) Practice

本文介绍了一个算法问题——计数家族树中各层级的叶子节点数量。输入包括节点总数及非叶子节点及其子节点信息，输出为从根节点开始各层级的叶子节点数。文章提供了完整的C++代码实现，通过递归方式处理森林结构，适用于解决类似计数问题。

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

注意：要考虑森林，刚开始没有考虑森林的情况，只考虑了一棵树导致一个点始终过不去。。。

#include <bits/stdc++.h>
using namespace std;
vector <int> v[105];
void fun(vector <int> d)
{
	vector <int> a;
	int cnt=0;
	int m=0;
	for(int i=0;i<d.size();i++)
	{
		m+=v[d[i]].size();
		for(int j=0;j<v[d[i]].size();j++)
		{
			if(v[v[d[i]][j]].size()==0)
				cnt++;
			else
				a.push_back(v[d[i]][j]);	
		}
	}
	if(cnt==m)
	{
		cout<<cnt;
		return;
	}
	cout<<cnt<<" ";
	fun(a);
}
int main()
{
	int n,m;
	cin>>n>>m;
	map <int,int> M;
	for(int i=0;i<m;i++)
	{
		int k,a,b;
		cin>>a>>k;
		for(int j=0;j<k;j++)
		{
			cin>>b;
			M[b]=1;
			v[a].push_back(b);
		}
	}
	vector <int> c;
	for(int j=1;j<=n;j++)
	{
		if(!M.count(j))
		{
			c.push_back(j);	
		}
	}
	int t=0;
	for(int j=0;j<c.size();j++)
	{
		if(v[c[j]].size()==0)
			t++;
	}
	if(t==c.size())
	{
		cout<<t;
		return 0;
	}
	else
	{
		cout<<t<<" ";
		fun(c);
	}	
	return 0;
}