PAT (Advanced Level) 1004 Counting Leaves (30分)

PAT (Advanced Level) 1004 Counting Leaves (30分)

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 <bits/stdc++.h>
using namespace std;

int n, m, maxdep;
vector<int> dep(105), num(105);
vector<vector<int>> graph(105);

void dfs(int u){
    for (auto v: graph[u]){
        dep[v] = dep[u] + 1;
        maxdep = max(maxdep, dep[v]);
        dfs(v);
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m;
    if (n == 0) {
        return 0;
    }
    for (int i = 0; i < m; i++){
        int K, u, v;
        cin >> u;
        cin >> K;
        for (int j = 0; j < K; j++){
            cin >> v;
            graph[u].push_back(v);
        }
    }
    dfs(1);
    for (int i = 1; i <=  n; i++){
        if (graph[i].size() == 0) {
            num[dep[i]]++;
        }
    }
    cout << num[0] ;
    for (int i = 1; i <= maxdep; i++){
        cout << " " << num[i];
    }
    cout << '\n';
    return 0;
}