1004 Counting Leaves (30 分)——甲级（dfs）

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

题目大意：给出一棵树，输出每一层的叶子节点的个数。
思路：用二维数组记录每个结点和这个结点的所有儿子结点，用cnt数组记录第depth层的叶子节点个数，然后从根节点开始dfs，每次dfs儿子结点时depth+1。

代码如下:

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int>v[100];
int cnt[100], height = -1;//height记录树的高度，后续输出结果会用到
void dfs(int index, int depth)
{
	if(!v[index].size())
	{
		cnt[depth]++;
		height = max( height, depth);
		return ;
	}
	for(int i=0; i < v[index].size(); i++)
		dfs( v[index][i], depth+1);
}
int main()
{
	int n, m, i;
	cin >> n;
	if(n)
	{
		cin >> m;
		while(m--)
		{
			int id, k;
			cin >> id >> k;
			while(k--)
			{
				int x;
				cin >> x;
				v[id].push_back(x);
			}
		}
		dfs(1, 0);
		cout << cnt[0];
		for(i = 1; i <= height; i++) cout << " " << cnt[i];
		cout << endl;
	}
	return 0;
}