1004 Counting Leaves

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
题意：给出一棵树，问每一层各有多少叶子结点
样例解释：
这棵树有两个结点，其中1个结点是非叶子结点
根结点01号有1个孩子，为02号结点。
输出0 1
（第一层有0个叶子结点，第二层有1个叶子结点）

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int N=110;
vector<int> G[N];//存放树
int leaf[N]={0};//存放每层的叶子结点个数
int max_h=1;//树的深度
void DFS(int index,int h){
	//index为当前遍历的结点编号，h为当前深度
	max_h=max(h,max_h);
	if(G[index].size()==0)
	{
		leaf[h]++;
		return;
	} 
	for(int i=0;i<G[index].size();i++)
	{
		DFS(G[index][i],h+1);
	}
} 
int main()
{
	int n,m,parent,child,k;
	scanf("%d%d", &n, &m);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&parent,&k);
		for(int j=0;j<k;j++)
		{
			scanf("%d", &child);
			G[parent].push_back(child);
		}
	}
	DFS(1,1);
	printf("%d",leaf[1]);
	for(int i=2;i<=max_h;i++)
	printf(" %d",leaf[i]);
	return 0;
}