1094. The Largest Generation

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最新推荐文章于 2024-07-16 16:03:51 发布

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分类专栏： PAT A C语言实现

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本文深入探讨了如何通过解析家族族谱数据，找到具有最大人口数量的群体及其所在层级。具体操作涉及输入数据格式、算法实现及输出结果解释。

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1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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<vector>
#include<stdio.h>
using namespace std;

int main()
{
 	//freopen("F://Temp/input.txt", "r", stdin);
	int n, m;
	cin>>n>>m;
	vector<int> v[100];
	while(m --)
	{
		int parent, k;
		cin>>parent>>k;
		for(int i = 0; i < k; i ++)
		{
			int child;
			cin>>child;
			v[parent].push_back(child);
		}
	}
	
	vector<int> v1, v2;
	int root = 1;
	v1.push_back(root);
	int max = 1, maxLevel = 1, curLevel = 1;
	while(1)
	{
		++ curLevel;
		if(v1.size() == 0)
			break;
		for(int i = 0; i < v1.size(); i ++)
		{
			for(int j = 0; j < v[v1[i]].size(); j ++)
				v2.push_back(v[v1[i]][j]);
		}
		
		if(v2.size() > max)
		{
			max = v2.size();
			maxLevel = curLevel;
		}
		v1 = v2;
		v2.clear();
	}
	cout<<max<<" "<<maxLevel<<endl;
	
	return 0;
}