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

思路
图的广搜(BFS)，或者可以看成树的层次遍历。仔细看 输入数据 会发现结构很像图的邻接表，所以用vector<vector<int>>模拟一个图的邻接表。所以:
1.根据数据构造图graph。
2.用队列bfs，记录下孩子最多的一层和最多孩子数。（层次的区分可以依靠插入一个label到队列区分，每当一个label出队时，表示这一层遍历完，并且这一层所有节点的孩子都刚好加入队列等待遍历，所以label出队的同时再插入一个label可以将孩子所在的层级与孩子的孩子所在的层级区分开来）
3.输出记录即可。
代码

#include<iostream>

#include<vector>

#include<queue>

//Need to be optimized

using namespace std;

int maxchild = ,level = ;

void bfs(int root,const vector<vector<int>>& g)

{

   int curlevel = ;

   maxchild = ;

   int countchild = ;

   int label = -;

   queue<int> q;

   q.push(root);

   q.push(label);

   while(!q.empty())

   {

       int f = q.front();

       q.pop();

       if( f == label)

       {

           ++curlevel;

           if(maxchild < countchild)

           {

              level = curlevel;

              maxchild = countchild;

           }

           countchild = ;

           if(q.empty())   //检查下label是不是最后一层的label

            break;

           q.push(label);

       }

       else

       {

           countchild++;

           for(int i = ;i < g[f].size();i++)

           {

               q.push(g[f][i]);

           }

       }

   }

}

int main()

{

   int N,M;

   while(cin >> N >> M)

   {

       //create graph

       vector<vector<int>> graph(N + );

       for(int i = ; i <=M;i++)

       {

           int node,childcount;

           cin >> node >> childcount;

           vector<int> childs(childcount);

           for(int j = ;j < childcount;j++)

            cin >> childs[j];

           graph[node] = childs;

       }

       bfs(,graph);

       cout << maxchild << " " << level;

   }

}