1004 Counting Leaves

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<cstdio>
#include<vector>
#include<queue>
using namespace std;

struct family
{
  int level;
  vector<int> child;
}families[110];

int total;
int non_leaf;

int ID;
int id;
int num_of_child;

int no_child[110];
int maxlevel = 0;

void level(int root)
{
  queue<family> Q;
  families[root].level = 1;
  Q.push(families[root]);

  while (!Q.empty())
  {
    family p = Q.front();
    Q.pop();
    for (int i = 0; i < p.child.size(); i++)
    {
      int chi = p.child[i];
      families[chi].level = p.level + 1;
      Q.push(families[chi]);
    }
  }
}

int main()
{
  scanf_s("%d %d", &total, &non_leaf);
  for (int i = 0; i < non_leaf; i++)
  {
    scanf_s("%d %d", &ID,&num_of_child);
    for (int j = 0; j < num_of_child; j++)
    {
      scanf_s("%d", &id);
      families[ID].child.push_back(id);
    }
  }

  level(1);

  for (int i = 1; i <= total; i++)
  {
    if (families[i].child.size() == 0)
    {
      no_child[families[i].level]++;
    }    
    if (families[i].level > maxlevel)
    {
      maxlevel = families[i].level;
    }
  }
  for (int i = 1; i <= maxlevel; i++)
  {
    printf("%d", no_child[i]);
    if (i < maxlevel)
    {
      printf(" ");
    }
    else
    {
      printf("\n");
    }
  }
}