题目地址

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

#include <iostream>
#include <vector>
#include<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=1010,inf=100000000;
int n,m;
vector<int>v[maxn];
int level[maxn]={0},max_dep=-1;
void dfs(int idx,int dep){
    if(dep>max_dep) max_dep=dep;
    level[dep]++;
    for(int i=0;i<v[idx].size();i++){
        dfs(v[idx][i],dep+1);
    }
}
int main(){
    int fa,chi,k;
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        cin>>fa>>k;
        for(int i=0;i<k;i++){
            cin>>chi;
            v[fa].push_back(chi);
        }
    }
    dfs(1,1);
    int ans1=-1,idx=-1;
    for(int i=0;i<=max_dep;i++){
        if(level[i]>ans1){
            ans1=level[i];idx=i;
        }
    }
    cout<<ans1<<" "<<idx;
}