PAT (Advanced Level) Practice 1004 Counting Leaves

分数 30
作者 CHEN, Yue
单位 浙江大学

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

题目翻译

背景：
族谱通常由系谱树来表示。你的工作是统计那些没有孩子的家庭成员总数。
输入格式：
每个输入文件包含一个测试用例。每一种情况都以包含以下内容：第一行有一个正整数N（0<N<100），它代表节点的总数，还有一个正整数M，代表不是叶节点的节点总数；然后接着有M行，每一行的格式都是 ID k ID[1] ID[2]…ID[K]，其中第一个ID是一个两位数，代表一个给定的非叶节点，K是它的子节点的数量，后面是它的全部子节点的两位数ID序列。为了简单起见，规定根节点的ID为01。
输出格式：
对于每个测试用例，你应该从根节点开始，计算每一层（同辈）的没有孩子的家庭成员总数。结果必须在一行中输出，用空格隔开，并且在每行的末尾不能有额外的空格。示例示例表示一棵只有2个节点的树，其中01是根节点，02是唯一的子节点。因此在根01层上，有0个叶节点;下一层，有1个叶节点。所以应该在一行中输出0 1。

代码

#include <iostream>
#include <string>
#include <stdlib.h>
#include <vector>using namespace std;int maxLevel=0,level[100]={0};
vector<int> tree[100];void dfs(int ID,int curLevel){if(maxLevel<curLevel)maxLevel=curLevel;if(tree[ID].size()>0){for(auto it:tree[ID]){dfs(it,curLevel+1);}}else level[curLevel]++;
}int main(){int n,m;cin>>n>>m;while(m--){int ID,k;cin>>ID>>k;for (int i = 0; i < k; i++) {int temp;cin >> temp;tree[ID].emplace_back(temp);}}dfs(1,1);for (int i = 1; i <= maxLevel; i++) {if (i!=1) cout<<" ";cout << level[i];}}