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 (<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
#include<iostream> #include<cstdio> #include<vector> using namespace std; typedef struct{ int data; vector<int>child; }node; node tree[20000]; int N, M, width[20000] = {0}, maxDepth = -1; void DFS(int s, int dp){ if(s > N) return; width[dp]++; if(dp > maxDepth){ maxDepth = dp; } for(int i = 0; i < tree[s].child.size(); i++){ DFS(tree[s].child[i], dp + 1); } } int main(){ scanf("%d%d", &N, &M); for(int i = 0; i < M; i++){ int id, K, temp; scanf("%d%d", &id, &K); tree[id].data = id; for(int j = 0; j < K; j++){ scanf("%d", &temp); tree[id].child.push_back(temp); } } DFS(1,1); int index = -1, max = -1; for(int i = 1; i <= maxDepth; i++){ if(width[i] > max){ max = width[i]; index = i; } } printf("%d %d", max, index); cin >> N; return 0; }