PTA The Largest Generation (25分)

释放无限光明的是人心，制造无边黑暗的也是人心，光明和黑暗交织着，厮杀着，这就是我们为之眷恋又万般无奈的人世间。

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 <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include<climits>//INT_MAX
//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=110;
vector<int> T[maxn];
int n,m,hs[maxn]={0},mx=0;
int id,k,tmp,ct;
void dfs(int root, int dh)
{
    hs[dh]++;
    if(T[root].size()==0)
        return ;
    for(int i=0;i<T[root].size();i++)
        dfs(T[root][i],dh+1);
}
int main()
{
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        cin>>id>>k;
        for(int i=0;i<k;i++)
        {
            cin>>tmp;
            T[id].push_back(tmp);
        }
    }
    dfs(1,1);
    for(int i=0;i<maxn;i++)
    {
        if(hs[i]>mx)
        {
            ct=i;
            mx=hs[i];
        }
    }
    cout<<mx<<' '<<ct<<endl;
    return 0;
}