PAT A1004 (30) [BFS， DFS，树的层序遍历]

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who
have no child.
Input
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.
Output
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 总结点数 ~ 非叶节点数
01 1 02 非叶结点ID ~ 孩子数 ~ 孩子ID
Sample Output
0 1第0层叶子结点数 ~ 。。。 ~ 第depth层叶子节点数

题⽬⼤意：给出⼀棵树，问每⼀层各有多少个叶⼦结点～
分析：用深度优先搜索，book[depth]：记录相应深度的叶结点数。v[ID].size()：对应ID孩子数。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> v[100];
int book[100], maxdepth = -1;

void dfs(int index, int depth) {cout<<index<<endl; 
	if(v[index].size() == 0) {
		book[depth]++;
		maxdepth = max(maxdepth, depth);
		return ;
	}
	for(int i = 0; i < v[index].size(); i++)
	dfs(v[index][i], depth + 1);
}

int main() {
	int n, m, k, node, c;
	scanf("%d %d", &n, &m);
	for(int i = 0; i < m; i++) {
		scanf("%d %d",&node, &k);
		for(int j = 0; j < k; j++) {
			scanf("%d", &c);
			v[node].push_back(c);
		}
	}
	dfs(1, 0);
	printf("%d", book[0]);
	for(int i = 1; i <= maxdepth; i++)
		printf(" %d", book[i]);
return 0;
}