PAT甲级 The Largest Generation (25)

题目描述
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.
输入描述:
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.
输出描述:
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.
输入例子:
23 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18
输出例子:
9 4
第一次自己编写的，牛客上排200，也很开心了。用的是层次遍历，额，目前也就这个可以比较号的掌握。这个题目比较清楚也就是二叉树遍历求每一次的节点数，每遍历一层就和上一层比较节点数，保留最大的。估计这个算笨办法。
#include<stdio.h>#include<iostream>#pragma  warning(disable:4786)#include<queue>#include<map>using namespace std;map<int,vector<int> > nodes;int m,n,levelest=0,levelnum=0;//levelest 最大节点数，levelnum 该层数void levelrange(){	queue<int> que;	if(nodes[1].size()==0) return;	int par = 1, child ; 	int count2=0,count1=1,level=1; //count2 当前层节点数，count1 上一层节点数	int lastLevelMember = 0;	bool first=true;	que.push(par);	while(!que.empty())	{		for(int i=0; i<count1; i++)		{			par= que.front();			que.pop();			for( int j=0; j < nodes[par].size();j++)			{				child = nodes[par][j];				que.push(child);				count2++;			}				}		level++;		lastLevelMember = count1;//上一层节点数，父辈		count1 = count2;		if(levelnum < count2) 		{	levelnum=count2; 			levelest = level;		}		count2 = 0;	}}int main(){	int pare,cnum,child;	scanf("%d%d",&m,&n);	for(int i=0; i<n; i++)	{		scanf("%d%d",&pare,&cnum);		for(int j=0; j<cnum; j++)		{			cin>>child;			nodes[pare].push_back(child);		}	}	levelrange();//层次遍历	printf("%d %d",levelnum,levelest);;}//