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
1 #include<iostream> 2 #include<string> 3 #include<stdlib.h> 4 #include<vector> 5 #define MaxNum 101 6 using namespace std; 7 typedef struct Node 8 { 9 int child = 0; 10 int level = -1; 11 int father; 12 }Tree[MaxNum]; 13 14 int main() 15 { 16 Tree tree; 17 int n, m; 18 cin >> n>> m; 19 for (int k = 0; k < m; k++) 20 { 21 int id,c,idj; 22 cin >> id >> c; 23 for (int j = 0; j < c; j++) 24 { 25 cin >> idj; 26 tree[id].child++; 27 tree[idj].father = id; 28 } 29 } 30 tree[1].father = 1; 31 tree[1].level = 0; 32 if (n == 1) 33 { 34 cout << "1" << endl; 35 return 0; 36 } 37 int flag = 1; 38 while (flag) 39 { 40 flag = 0; 41 for (int i = 1; i <=n; i++) 42 { 43 if (tree[tree[i].father].level != -1 && tree[i].level == -1)tree[i].level = tree[tree[i].father].level + 1; 44 else if (tree[tree[i].father].level == -1) 45 flag = 1; 46 } 47 } 48 int Level[MaxNum] = { 0 }; 49 int MaxLevel = -1; 50 for (int i =1; i <=n; i++) 51 { 52 if (tree[i].child== 0)Level[tree[i].level]++; 53 MaxLevel = MaxLevel > tree[i].level ? MaxLevel : tree[i].level; 54 } 55 cout << Level[0]; 56 for (int i = 1; i <= MaxLevel; i++) 57 cout << " " << Level[i]; 58 return 0; 59 }