【PAT】1094. The Largest Generation (25)【树的层次遍历】

题目介绍

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.

翻译：一个家族的层次总是被描述为一棵家谱树，所有同一层次的节点都有相同的辈分。你的任务是找出人数最多的辈分。

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.

翻译：每个输入文件包含一组测试数据。每组测试数据开头为两个正整数N(<100)，N表示书上的家庭成员的总人数(假设所有人员都被编号为1到N),M(<N)表示有孩子的家庭成员的人数。接下来M行，每一行包括一个家族成员的信息，按照以下格式：
ID K ID[1] ID[2] … ID[K]
ID为一个两位数字表示一个家庭成员，K(>0)为他/她的孩子数，接下来为一列他/她孩子的两位编号。为了计算简便，让我们固定根ID为1。一行内所有数字用空格隔开。

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.

翻译：对于每组输入数据，输出一行最大人数和对应的辈分。假设该辈分是唯一的，且根的辈分定义为1。

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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

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Sample Output:

9 4

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解题思路

这道题是一道典型的层次遍历题，我为了不建树，开了个数组来记录每个节点的父节点，再遍历一遍每个节点的层级，计算每个层级的节点数即可，详见代码。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#define INF 99999999
#define bug puts("Hello\n")
using namespace std;
int M,N;
int Parent[110]; //用于保存节点的父节点 
int Step[110];	 //用于保存节点的层级
int ccount[110]; //用于保存层级的节点数 
int find(int node){
	if(Step[node]);
	else if(node==1)Step[node]=1;
	else Step[node]=find(Parent[node])+1;
	return Step[node];
}
int main(){
	scanf("%d%d",&M,&N);
	int ID,K;
	for(int i=0;i<N;i++){
		scanf("%d %d",&ID,&K);
		int a;
		for(int j=0;j<K;j++){
			scanf("%d",&a);
			Parent[a]=ID;
		}
	}
	for(int i=1;i<=M;i++)find(i);
	for(int i=1;i<=M;i++)ccount[Step[i]]++;
	int ansStep=0,ansCount=0;
	for(int i=1;i<=M;i++){
		if(ccount[i]>ansStep){
			ansStep=i;
			ansCount=ccount[i];
		}
	} 
	printf("%d %d\n",ansCount,ansStep);
	return 0;
}