PAT (Advanced Level) Practice 1004 Counting Leaves (30分) （普通树的DFS）

1.题目

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

2.题目分析

题目的数据结构是一棵普通的树型结构，使用DFS遍历，遇到叶子节点记录即可

3.代码

#include<iostream>
#include<vector>
using namespace std;
vector<int>list[110];
int n, m;
//int visited[110];
int deep[110];//对应数某层的叶子节点个数
int level=0;//树的层数
void DFS(int head, int times)
{
	//visited[head] = 1;//这个基于vector数组的DFS不用visited维护，因为是按照顺序访问，彼此无交叉，毕竟是树而不是严格意义上的图
    if (level < times)level = times;//记录树的总层数
	if (list[head].size() == 0)deep[times]++;//是叶子节点就在相应位置的个数加一
	else
		for (int i = 0; i<list[head].size(); i++)
		{
			//if (visited[list[head][i]] == 0)
			//{
				times++;
				DFS(list[head][i], times);
				times--;
			//}
		}
}
int main()
{
	int k, a, b;
	scanf("%d", &n);
	if (n == 0)return 0;
	scanf("%d", &m);
	for (int i = 1; i <= m; i++)
	{
		scanf("%d %d", &a, &k);
		for (int j = 0; j<k; j++)
		{
			scanf("%d", &b);
			list[a].push_back(b);
		}
	}
	DFS(1, 0);
	for (int i = 0; i<level+1; i++)
		printf("%s%d", i == 0 ? "" : " ", deep[i]);

}