PAT (Advanced Level) Practice 1094 The Largest Generation （25 分）-优快云博客

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本文探讨了如何通过编程解决家族树中寻找拥有最多成员的一代的问题。输入包含家庭成员总数及部分成员的孩子信息，输出最大代数的人口数量及其对应的代数。使用DFS遍历算法，累计每一代的成员数量，并找出人数最多的一代。

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编程题

1094 The Largest Generation （25 分）

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

题目大意：

给出根节点和每个非叶节点的子节点，以此建树，求节点数量最多的层的节点个数。

解题思路：

DFS，遍历每个节点，所在的层数的节点个数累计。同时记录最大层数。

从第1层到最大层数间，找到个数最多的层数，输出。

注意事项：记录最大层数有助于找个数最多层的时候节省时间。

源代码1：

#include<iostream>
#include<stdlib.h>
#include<vector>
using namespace std;
struct node
{
	vector<int>child;
};
vector<node>v;
int cnt[101] = { 0 };
void dfs(int j, int level)
{
	cnt[level]++;
	for (int i = 0; i < v[j].child.size(); i++)
	{
		dfs(v[j].child[i], level + 1);
	}
	return;
}
int main()
{
	int n,m,temp,j,k;
	scanf("%d %d", &n,&m);
	v.resize(n+1);
	for (int i = 0; i < m; i++)
	{
		scanf("%d %d", &temp,&k);
		v[temp].child.resize(k);
		for (j = 0; j < k; j++)
			scanf("%d", &v[temp].child[j]);
	}
	dfs(01, 1);
	int maxlevel = 1;
	for (int i = 1; i <= n; i++)
		if (cnt[i] > cnt[maxlevel])maxlevel = i;
	printf("%d %d", cnt[maxlevel],maxlevel);
	system("pause");
	return 0;
}