【PAT甲级】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

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

【翻译】：输入N,M，分别代表树的节点数和非叶子节点总数，接下来的M行输入非叶子节点ID（两位数），k，孩子数，ID[1]......ID[k]，孩子节点的ID，输入以0结尾，输出从根节点开始每一层孩子节点数，数据间有空格，行尾没有空格。

又是参考柳神的，确实很easy，就是用dfs，首先用vector构建数，然后用深搜搜每一层就好了。 

#include 
#include 
#include 
#define MAX 1000
using namespace std;
int ans[MAX];
int maxdep = -1;
vector v[MAX];
void dfs(int index,int depth){
	if(v[index].size() == 0){
		ans[depth]++;
		maxdep = max(depth, maxdep);
		return;
	}
	for(int i = 0; i < v[index].size(); i++){
		dfs(v[index][i], depth + 1);
	}
}
int main(){
	int n,m,k,c,id;//n是树的节点数，m树中无叶子节点的节点数 
	cin>>n>>m;
	for(int i = 0; i < m; i++){
		cin>>id>>k;
		for(int j = 0; j < k; j++){
			cin>>c;
			v[id].push_back(c);
		}
	}
	dfs(1,0);
	cout<