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[];
int N, M, width[] = {}, maxDepth = -;
void DFS(int s, int dp){
if(s > N)
return;
width[dp]++;
if(dp > maxDepth){
maxDepth = dp;
}
for(int i = ; i < tree[s].child.size(); i++){
DFS(tree[s].child[i], dp + );
}
}
int main(){
scanf("%d%d", &N, &M);
for(int i = ; i < M; i++){
int id, K, temp;
scanf("%d%d", &id, &K);
tree[id].data = id;
for(int j = ; j < K; j++){
scanf("%d", &temp);
tree[id].child.push_back(temp);
}
}
DFS(,);
int index = -, max = -;
for(int i = ; i <= maxDepth; i++){
if(width[i] > max){
max = width[i];
index = i;
}
}
printf("%d %d", max, index);
cin >> N;
return ;
}
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