A1004 Counting Leaves (30 分)PAT甲级（C++）【DFS】

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

代码如下：

#include
#include
#include
using namespace std;
vector v[100];//存放每一个节点，vector的每一个元素都是一个容器，存放一个节点的所有儿子
int book[100], maxdepth = -1;//book中存放每一层的叶子结点个数
//ID k ID[]...
void dfs(int index, int depth)
{
	if (v[index].size() == 0) {//容器中没有元素，即该节点没有儿子，所以为叶子结点
		book[depth]++;//叶子结点个数增加
		maxdepth = max(maxdepth, depth);//判断深度是否要增加
		return;
	}
	for (int i = 0; i < v[index].size(); i++)
	{
		dfs(v[index][i], depth + 1);
	}
}
int main()
{
	int n, m, k, node, c;//n为树中的节点数量，m为非叶子节点的数量
	//cin >> n >> m;
	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);//将该节点加入儿子节点容器中
		}
	}
	dfs(1, 0);
	printf("%d", book[0]);
	for (int i = 1; i <= maxdepth; i++)
	{
		printf(" %d", book[i]);
	}
	return 0;
}

运行结果如下：