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

 1 #include <iostream>
 2 #include <queue>
 3 #include <vector>
 4 using namespace std;
 5 int N, M, maxN = 1, resL = 1, root = 1, level[105] = { 0 }, manN[105] = { 0 };
 6 vector<int>man[105];
 7 void BFS()
 8 {
 9     queue<int>q;
10     q.push(root);
11     level[root] = 1;
12     manN[level[root]]++;
13     while (!q.empty())
14     {
15         root = q.front();
16         q.pop();
17         int temp = 0;
18         for (auto v : man[root])
19         {
20             level[v] = level[root] + 1;
21             manN[level[v]]++;//记录每一层的人数
22             if (man[v].size() > 0)
23                 q.push(v);
24         }
25     }    
26 }
27 
28 void DFS(int s,int l)
29 {
30     manN[l]++;//l层的人数
31     for (auto v : man[s])
32         DFS(v, l + 1);
33 }
34 
35 int main()
36 {
37     cin >> N >> M;
38     for (int i = 0; i < M; ++i)
39     {
40         int a, b, k;
41         cin >> a >> k;
42         for (int j = 0; j < k; ++j)
43         {
44             cin >> b;
45             man[a].push_back(b);
46         }
47     }
48     //BFS();
49     DFS(1, 1);
50     for (int i = 1; i <= N; ++i)
51     {
52         if (maxN < manN[i])
53         {
54             maxN = manN[i];
55             resL = i;
56         }
57     }    
58     cout << maxN << " " << resL << endl;
59     return 0;
60 }

PAT甲级——A1094 The Largest Generation

原文：https://www.cnblogs.com/zzw1024/p/11350268.html