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：
　　给一棵树，求这棵树中每一层中，没有子节点的节点个数。第一行输入两个数n,m；分别表示这棵树节点和叶子节点个数，第二行依次输入节点编号和这个节点子节点个数

题解：
　　用vector数组保存每个节点的儿子节点，然后用DFS依次遍历每个深度的每个节点，数组记录每层子节点数为0的节点即可

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#define MAX 1000000
using namespace std;
int num[1005],vis[10005];
vector<int>v[1005];
int cnt=0;
void dfs(int root,int dep)
{
    vis[root]=1;
    if(v[root].size()==0)
    {
        num[dep]++;
        cnt=cnt<dep?dep:cnt;//记录最大深度
    }
    else
    {
        for(int i=0;i<v[root].size();i++)//遍历儿子节点
        {
            if(vis[v[root][i]]==0)
                dfs(v[root][i],dep+1);
        }
    }
}
int main()
{
    int n,m,id,k;
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        cin>>id>>k;
        for(int i=0;i<k;i++)
        {
            int chi;//儿子节点
            cin>>chi;
            v[id].push_back(chi);
        }
    }
    dfs(1,0);
    for(int i=0;i<=cnt;i++)
    {
        if(i==0)
            cout<<num[i];
        else
            cout<<‘ ‘<<num[i];
    }
    cout<<endl;
    return 0;
}