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

#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;

vector<int> v[110];
int ans[110] = {0};
int maxdepth = -1;
void DFS(int root,int depth){
	if(v[root].size()==0){
		ans[depth]++;
		maxdepth = max(maxdepth, depth);
	}
	for(int i=0;i<v[root].size();i++){
		DFS(v[root][i],depth+1);
	}
}
int main(){
	int n,m,a,b,c;
	cin>>n>>m;
	while(m--){
		cin>>a>>b;
		for(int i=0;i<b;i++){
			cin>>c;
			v[a].push_back(c);
		}
	}
	DFS(1,0);
	bool flag = true;
	for(int i=0;i<maxdepth+1;i++){
		printf("%s%d",flag==true?"":" ",ans[i]);
		flag = false;
	}		
	return 0;
}