PAT甲级 1004 Counting Leaves (30 分)

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该程序实现了一个用于计算家族树中每个级别无子节点成员数量的功能。输入是一个家族树的非叶节点及其子节点信息，输出是从根节点开始各级别的无子节点人数。给定的示例输入是一个有两个节点的树，其中01是根节点，02是其唯一子节点。程序首先进行深度优先搜索遍历，然后输出每个级别的无子节点数量。对于给定的示例，输出为01，表示根节点级别有0个无子节点，下一级别有1个无子节点（即02）。

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

#include <iostream>
#include <vector>

using namespace std;
vector<int> v[105];
int cnt[105];
int maxDepth = -1;

void dfs(int node, int depth) {
    if (v[node].empty()) {
        cnt[depth]++;
        maxDepth = max(maxDepth, depth);
        return;
    }
    for (int i = 0; i < v[node].size(); ++i) {
        dfs(v[node][i], depth + 1);
    }
}

int main() {
    int n, m, k, g, h;
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        cin >> k >> g;
        for (int j = 0; j < g; ++j) {
            cin >> h;
            v[k].push_back(h);
        }
    }
    dfs(1, 0);
    cout << cnt[0];
    for (int i = 1; i <= maxDepth; ++i) {
        cout << " " << cnt[i];
    }
    return 0;
}