PAT甲级 The Largest Generation (25)_the largest generation (25)柳婼-优快云博客

本文介绍了一种解决家族树中寻找人口最多一代的问题的方法。通过层次遍历算法，逐层计算各代人数并记录最大值及其对应层级，最终输出答案。

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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 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

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);;
}
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