1094 The Largest Generation （25 分）

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>
#include<map>
using namespace std;

struct Node{
	vector<int> child;
}node[110];

map<int,int> mp;

void BFS(int root,int width){
	mp[width]++;
	if(node[root].child.size()==0){
		return;
	}else{
		for(int i=0;i<node[root].child.size();i++){
			BFS(node[root].child[i],width+1);
		}
	}
}

int main(){
	int n,m,a,b,c;
	cin>>n>>m;
	while(m--){
		cin>>a>>b;
		for(int i=0;i<b;i++){
			cin>>c;
			node[a].child.push_back(c);
		}		 
	}
	BFS(1,0);
	int max1 = 0,max2=0;
	for(map<int,int>::iterator it=mp.begin();it!=mp.end();it++){
		if(max2<it->second){
			max2 = it->second;
			max1 = it->first;
		}
	}
	cout<<max2<<" "<<max1+1;
	return 0;
}