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.
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.
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.
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
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