PAT.A1094 The Largest Generation

 

void bfs() {
	int level = 2, last = 1, first = 0, rear = 0, num = 0;
	queue<int> q;
	q.push(1);
	//l[1] = 1;
	rear++;
	lnum[1] = 1;
	while (!q.empty()) {
		int f = q.front();
		q.pop();
		first++;
	//	lnum[l[f]]++;
		for (int i = 0; i < child[f].size(); i++) {
			q.push(child[f][i]);
			rear++;
			num++;
			//l[child[f][i]] = l[f] + 1;
			}
		if (first == last) {
			last = rear;
			lnum[level++] = num;
			num = 0;
		}
		}
	for (int i = 1; i < level; i++) {
		if (lnum[i] > maxnu) {
			maxnu = lnum[i];
			maxl = i;
		}
		
	}
	//printf("\n\n");
}

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>																							
#include<vector>
#include<queue>
#include<cstring>
#include<stack>
using namespace std;
const int maxn = 110;
int n, m, l[maxn] = { 0 }, maxnu = 0, maxl = 0,lnum[maxn] = { 0 };
vector<int> child[maxn];
void bfs() {
	queue<int> q;
	q.push(1);
	l[1] = 1;
	while (!q.empty()) {
		int f = q.front();
		q.pop();
		lnum[l[f]]++;
		for (int i = 0; i < child[f].size(); i++) {
			q.push(child[f][i]);
			l[child[f][i]] = l[f] + 1;
			}
		}
	for (int i = 0; i < maxn; i++) {
		if (lnum[i] > maxnu) {
			maxnu = lnum[i];
			maxl = i;
		}
	}
}
void dfs(int root,int depth) {
	l[depth]++;
	if (l[depth] > maxnu) {
		maxnu = l[depth];
		maxl = depth;
	}
	if (child[root].size() == 0) return;
	for (int i = 0; i < child[root].size(); i++)
		dfs(child[root][i], depth + 1);
}
int main() {
	scanf("%d%d", &n, &m);
	memset(l, 0, sizeof(l));
	memset(lnum, 0, sizeof(lnum));
	for (int i = 0; i < m; i++) {
		int id,k;
		scanf("%d%d", &id,&k);
		for (int j = 0; j < k; j++) {
			int tmp;
			scanf("%d", &tmp);
			child[id].push_back(tmp);
		}
	}
	//dfs(1,1);
	bfs();
	printf("%d %d", maxnu, maxl);
	return 0;
}

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