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;
}