PAT:1004 Counting Leaves (30分)（dfs遍历）-CSDN博客

本文链接：https://blog.csdn.net/stackempty2/article/details/108983867

该博客介绍了一道PAT算法题目，涉及家族树的表示和遍历。题目要求计算从根节点开始，每一层没有孩子的家庭成员数量。输入包含一棵不超过100个节点的树，输出各层级的叶子节点数。博主提供了使用深度优先搜索（DFS）解决此问题的思路，通过遍历树结构记录每层的叶子节点数目。

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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

这道题还是思路相当清楚地，用book来存放深度上的叶子结点，dfs往下一直走就OK了

#include<bits/stdc++.h>
using namespace std;
vector<int>v[100010];
int book[10010];
int maxlevel;
int level;
void dfs(int root,int level)
{
    if(v[root].size()==0){
        book[level]++;
        maxlevel=max(maxlevel,level);
        return ;
    }
    for(int i=0;i<v[root].size();i++){
        dfs(v[root][i],level+1);
    }
}
int main()
{
    int n,m;
    cin>>n>>m;
    int a,b,k;
    for(int i=0;i<m;i++){
        cin>>a>>k;
        for(int j=0;j<k;j++){
            cin>>b;
            v[a].push_back(b);
        }
    }
    dfs(1,0);
    for(int i=0;i<=maxlevel;i++){
        if(i==0) printf("%d",book[i]);
        else printf(" %d",book[i]);
    }
}