A1094(The Largest Generation)

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

//The Largest Generation
//多叉树的遍历
#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
//结点结构体
struct Node
{
    string ID;    //编号
    int childNum; //孩子的数目
    vector<Node> Children;
};

void count(vector<Node> v)
{
    //层序遍历多叉数
    vector<int> level;
    int maxCount = 1;
    int t = 0;
    queue<Node> Q;
    Q.push(v[1]); //root push
    t++;
    while (!Q.empty())
    {
        int size = Q.size();

        Node head = Q.front();
        for (int i = 0; i < head.childNum; ++i)
        {
            //输入子代
            string id = head.Children[i].ID;
            int temp = id[0] == '0' ? id[1] - '0' : atoi(id.c_str());
            Q.push(v[temp]);
        }
        Q.pop();
        t--;
        if (t == 0)
        {
            maxCount = max(size, maxCount);
            level.push_back(maxCount);
            t = size;
        }
    }
    cout << maxCount << " ";
    for (t = 0; t < level.size(); ++t)
    {
        if (level[t] == maxCount)
            break;
    }
    cout << ++t;
}
int main(int argc, char const *argv[])
{
    /* code */
    int nodes = 0;
    int row = 0;
    scanf("%d%d", &nodes, &row);
    vector<Node> arr(nodes + 1);
    for (int i = 1; i <= 23; ++i)
    {
        arr[i].ID = (i < 10) ? "0" + to_string(i) : to_string(i);
    }
    string in;
    int n;
    for (int i = 0; i < row; ++i)
    {
        cin >> in;
        int p = in[0] == '0' ? in[1] - '0' : atoi(in.c_str());
        cin >> n;
        for (int j = 0; j < n; ++j)
        {
            cin >> in;
            Node son;
            son.ID = in;
            arr[p].Children.push_back(son);
        }
        arr[p].childNum = arr[p].Children.size();
    }
    count(arr);

    return 0;
}