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, the number of nodes in a tree, and M (<), 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
 
题目分析 ：第一行给了你2个数字，一个代表的是总节点数，一个代表的是叶子节点数 之后的几行 父亲节点与子节点，要求算出每一层叶子节点的数量
参考了别人的答案https://blog.csdn.net/qq_37613112/article/details/90577948
就是先将所有节点都先录入，然后对所有节点遍历，当它父亲节点的层数确定好后，它自己的层数也就能确定了

 #include<iostream>
 #include<string>
 #include<stdlib.h>
 #include<vector>
 #define MaxNum 101
 using namespace std;
 typedef struct Node
 {
     int child = ;
     int level = -;
     int father;
 }Tree[MaxNum];
 
 int main()
 {
     Tree tree;
     int n, m;
     cin >> n>> m;
     for (int k = ; k < m; k++)
     {
         int id,c,idj;
         cin >> id >> c;
         for (int j = ; j < c; j++)
         {
             cin >> idj;
             tree[id].child++;
             tree[idj].father = id;
         }
     }
     tree[].father = ;
     tree[].level = ;
     if (n == )
     {
         cout << "" << endl;
         return ;
     }
     int flag = ;
     while (flag)
     {
         flag = ;
         for (int i = ; i <=n; i++)
         {
             if (tree[tree[i].father].level != - && tree[i].level == -)tree[i].level = tree[tree[i].father].level + ;
             else if (tree[tree[i].father].level == -)
                 flag = ;
         }
     }
     int Level[MaxNum] = {  };
     int MaxLevel = -;
     for (int i =; i <=n; i++)
     {
         if (tree[i].child== )Level[tree[i].level]++;
         MaxLevel = MaxLevel > tree[i].level ? MaxLevel : tree[i].level;
     }
     cout << Level[];
     for (int i = ; i <= MaxLevel; i++)
         cout << " " << Level[i];
     return ;
 }