## Problem #

Source: [PAT ]

### Description #

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`

## Solution #

## Code #

```
#include <string>
#include <iostream>
#include <vector>
#include <queue>
#include <string>
#include <math.h>
#define max_size 101
using namespace std;
struct node
{
int deepth;
vector<int> sons;
node()
{
deepth = 1;
}
};
int number[max_size] = {0}, n, m;
bool isRoot[max_size];
node tree[max_size];
void dfs(int root)
{
++number[tree[root].deepth];
for (int i = 0; i < tree[root].sons.size(); i++)
{
int item = tree[root].sons[i];
tree[item].deepth = tree[root].deepth + 1;
dfs(item);
}
}
int main()
{
cin >> n >> m;
fill(isRoot+1, isRoot + n+1, true);
for (int i = 0; i < m; i++)
{
string sm, sx;
int m, t, x;
cin >> sm >> t;
m = (sm[0] - '0') * 10 + sm[1] - '0';
for (int j = 0; j < t; j++)
{
cin >> sx;
x = (sx[0] - '0') * 10 + sx[1] - '0';
tree[m].sons.push_back(x);
isRoot[x] = false;
}
}
int root;
for (root =1; root <= n; root++)
{
if (isRoot[root])
break;
}
dfs(root);
int u = 0;
root = 0;
for (int i = 1; i <= n; i++)
{
if (root < number[i])
{
root = number[i];
u = i;
}
}
cout << root << " " << u;
return 0;
}
```