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

解题思路：

这道题是给你一棵树，让你求出每一层的叶子节点个数，我们只需要使用DFS或者BFS遍历这棵树，然后分别记录每一层叶子节点数就行了，很简单。

#include 
#include 
#include 
#include 
using namespace std;

const int N = 110;

vector G[N];    //用来记录树的信息
int max_h = 1;		 //记录树的最大深度
int leaf[N] = {0};	 //每一层中叶子节点的个数

void DFS(int index, int h){
	max_h = max(max_h, h);
	if(G[index].size() == 0){		//是叶子结点 
		leaf[h]++;
		return;
	}
	for(int i=0; i