本文同步发布在CSDN：https://blog.csdn.net/weixin_44385565/article/details/93311728
A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.
Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (<) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:
ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a family member, K (>) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.
Sample Input:
23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18
Sample Output:
9 4题目大意：一个家族里面的人的ID编号是从01到N，每个人有若干个孩子，形成一棵族谱树，同一层的人是同辈关系，找出人数最多的那一层并且输出该层的人数。
思路：层序遍历（BFS思想）标记每个人所在的层数，另开一个数组 ans[level] = num 用于映射层数和人数的关系，树遍历完成后再遍历ans数组就能找到人数最多的那一层。
1 #include <iostream> 2 #include <vector> 3 #include <queue> 4 using namespace std; 5 6 struct node { 7 int level = 0; 8 vector<int> child; 9 }; 10 vector<node> tree; 11 vector<int> ans; 12 void getLevel(); 13 int main() 14 { 15 int N, M, ID, K; 16 scanf("%d%d", &N, &M); 17 tree.resize(N + 1); 18 ans.resize(N + 1, 0); 19 for (int i = 0; i < M; i++) { 20 scanf("%d%d", &ID, &K); 21 tree[ID].child.resize(K); 22 for (int j = 0; j < K; j++) 23 scanf("%d", &tree[ID].child[j]); 24 } 25 getLevel(); 26 int level = 0, num = 0; 27 for (int i = 0; i < ans.size(); i++) { 28 if (num < ans[i]) { 29 num = ans[i]; 30 level = i; 31 } 32 } 33 printf("%d %d ", num, level); 34 return 0; 35 } 36 37 void getLevel() { 38 int ID = 1; 39 queue<int> Q; 40 tree[ID].level = 1; 41 Q.push(ID); 42 while (!Q.empty()) { 43 ID = Q.front(); 44 ans[tree[ID].level]++; 45 Q.pop(); 46 int childID; 47 for (int i = 0; i < tree[ID].child.size(); i++) { 48 childID = tree[ID].child[i]; 49 tree[childID].level = tree[ID].level + 1; 50 Q.push(childID); 51 } 52 } 53 }