PAT 1094. The Largest Generation (层级遍历）

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

分析

用邻接表建树，层级遍历更新数据,用last,tail更新层级

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
int main(){
	int n,m,father,son,num,last=1,tail,cnt=0,max=-1,level,tag=1;
	cin>>n>>m;
	vector<int> family[n+1];
	for(int i=0;i<m;i++){
		cin>>father>>num;
		for(int j=0;j<num;j++){
			cin>>son;
			family[father].push_back(son);
		}
	}
	queue<int> q;
	q.push(1);
	while(!q.empty()){
		int t=q.front();
		q.pop();
		cnt++;
		for(int i=0;i<family[t].size();i++){
			q.push(family[t][i]);
			tail=family[t][i];
		}
		if(t==last){
			last=tail;
			if(cnt>max){
				max=cnt;
				level=tag;
			}
			tag++;
			cnt=0;
		}
	}
	cout<<max<<" "<<level;
	return 0;
}