1094 The Largest Generation (25 分)
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 (<100) 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 (<N) 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 (>0) 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：给定结点数和非叶结点的个数，以及非叶结点的子结点。求结点数最多的一层，设根结点的层号为1
分析：用DFS模板，传入结点编号和层号，每遍历一个结点，该结点对应的level++。最后遍历level数组，找到拥有最多结点数的层数

AC代码

#include<bits/stdc++.h>
using namespace std;
int n, m,level[110],maxNode,maxIndex;
vector<vector<int>> node;
void DFS(int index,int height){
  level[height]++;//每遍历一个结点时，对应层数的level++
  for (int i = 0; i < node[index].size();i++)
    DFS(node[index][i], height + 1);
}
int main(){
  node.resize(110);
  cin >> n >> m;
  int a, b, c;
  for (int i = 0; i < m;i++){
    cin >> a >> b;
    for (int j = 0; j < b;j++){
      cin >> c;
      node[a].push_back(c);
    }
  }
  DFS(1, 1);//根结点为1，根结点的层数默认为1
  for (int i = 1; i <= n;i++){
    if(level[i]>maxNode){
      maxNode = level[i];
      maxIndex = i;
    }
  }
  printf("%d %d", maxNode, maxIndex);

  return 0;
}