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<algorithm>
#include<vector>
#define MAX 100
using namespace std;
int maxdepth = -1;
vector<int> v[MAX];
int childs[MAX] = {};
void dfs(int index,int depth){
if(v[index].size() == 0){
childs[depth]++;
maxdepth = max(maxdepth,depth);
return;
}
for(int i = 0; i < v[index].size(); i++){
dfs(v[index][i], depth + 1);
}
}
int main(){
int m,n;
int id,k,id_n;
cin >> n >> m;
for(int i = 0; i < m; i++){
cin >> id >> k;
for(int j = 0; j < k; j++){
cin >>id_n;
v[id].push_back(id_n);
}
}
dfs(1, 0);
cout << childs[0];
for(int i = 1; i <= maxdepth; i++){
cout << " " << childs[i];
}
return 0;
}