1004 Counting Leaves (30 分)

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<vector>
#include<algorithm>
using namespace std;
vector<int> id[100];
int book[100];
int maxdepth = -1;
void dfs(int ID, int depth) {
	if (id[ID].size() == 0) {
		book[depth]++;
		maxdepth = max(maxdepth, depth);
		return;
	}
	for (int i = 0; i < id[ID].size(); i++)
		dfs(id[ID][i], depth + 1);
}
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int ID, K;
		cin >> ID >> K;
		for (int j = 0; j < K; j++) {
			int temp;
			cin >> temp;
			id[ID].push_back(temp);
		}
	}
	dfs(1, 0);
	cout << book[0];
	for (int k = 1; k <= maxdepth; k++)
		cout << " " << book[k];
	return 0;
}