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 IDto 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
  
 1 #include <stdio.h>
 2 #include <vector>
 3 #include <queue>
 4 using namespace std;
 5 const int maxn=110;
 6 vector<int> fa[maxn];
 7 int main(){
 8   int n,m;
 9   scanf("%d %d",&n,&m);
10   for(int i=1;i<=m;i++){
11     int root,k;
12     scanf("%d %d",&root,&k);
13     for(int j=0;j<k;j++){
14       int ch;
15       scanf("%d",&ch);
16       fa[root].push_back(ch);
17     }
18   }
19   queue<int> q;
20   q.push(1);
21   int maxm=1,lvl=1,max_l=1;
22   while(!q.empty()){
23     queue<int> child;
24     int num=0;
25     while(!q.empty()){
26       int now = q.front();
27       q.pop();
28       for(int i=0;i<fa[now].size();i++){
29         child.push(fa[now][i]);
30         num++;
31       }
32     }
33     lvl++;
34     if(num>maxm){
35       maxm=num;
36       max_l=lvl;
37     }
38     while(!child.empty()){
39       q.push(child.front());
40       child.pop();
41     }
42   }
43   printf("%d %d",maxm,max_l);
44 }
View Code

注意点：统计每层个数，用两个队列实现，同时统计个数和层数，一层全遍历完，再把下一层加入到队列中去