PAT甲级——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 (<) 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 (<) 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 (>) 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 <queue>
 #include <vector>
 using namespace std;
 int N, M, maxN = , resL = , root = , level[] = {  }, manN[] = {  };
 vector<int>man[];
 void BFS()
 {
     queue<int>q;
     q.push(root);
     level[root] = ;
     manN[level[root]]++;
     while (!q.empty())
     {
         root = q.front();
         q.pop();
         int temp = ;
         for (auto v : man[root])
         {
             level[v] = level[root] + ;
             manN[level[v]]++;//记录每一层的人数
             if (man[v].size() > )
                 q.push(v);
         }
     }
 }

 void DFS(int s,int l)
 {
     manN[l]++;//l层的人数
     for (auto v : man[s])
         DFS(v, l + );
 }

 int main()
 {
     cin >> N >> M;
     for (int i = ; i < M; ++i)
     {
         int a, b, k;
         cin >> a >> k;
         for (int j = ; j < k; ++j)
         {
             cin >> b;
             man[a].push_back(b);
         }
     }
     //BFS();
     DFS(, );
     for (int i = ; i <= N; ++i)
     {
         if (maxN < manN[i])
         {
             maxN = manN[i];
             resL = i;
         }
     }
     cout << maxN << " " << resL << endl;
     return ;
 }