1004 Counting Leaves (30分)（BFS DFS）

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
作者
CHEN, Yue
单位
浙江大学
代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB
题目大意：给出一棵树，问每一层各有多少个叶子结点～
分析：
用二维数组存储每个节点的孩子节点，再利用深度优先搜索计算出每一层叶子结点的个数。

# include 
using namespace std;
vector children[102];
int leaf[102];
int maxfloor=0;
void dfs(int p,int floor){
	 
	 maxfloor=max(maxfloor,floor);    
	 long l=children[p].size();
	 if(l==0){      
		 leaf[floor]++;
		  return;          /*当为叶结点时回溯到该节点的父节点*/
	 }
	    floor++;
	  for(int i=0;i