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

Process

模拟n叉树的广度优先搜索即可。

Code

#include<bits/stdc++.h>
using namespace std;
#define maxn 100010
int num[maxn];
struct node {
	int high = 0;
	vector<int> child;
}link[maxn];
int n, m;
void bfs() {
	queue<int >q;
	q.push(1);
	link[1].high = 1;
	while (!q.empty()) {
		int top = q.front();
		q.pop();
		for (int i = 0; i < link[top].child.size(); i++)
		{
			int temp = link[top].child[i];
			q.push(temp);
			link[temp].high = link[top].high + 1;
		}
	}
}
int main()
{
	cin >> n >> m;
	for (int i = 0; i < m; i++)
	{
		int temp, tempn;
		cin >> temp >> tempn;
		for (int j = 0; j < tempn; j++) {//孩子节点
			int t;
			cin >> t;
			link[temp].child.push_back(t);
		}
	}
	bfs();//广度优先
	int maxx = -1, maxm = -1;
	for (int i = 1; i <= n; i++)
		num[link[i].high]++;
	for (int i = 1; i <= n; i++)
		if (maxx < num[i])
		{
			maxx = num[i];
			maxm = i;
		}
	cout << maxx << " " << maxm;
	return 0;
}