1094 The Largest Generation (25 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/Checkmate2000/article/details/122826045

该程序用于解决家族树中最大人口世代的问题。输入包含家庭成员总数和有孩子的成员数量，然后列出每个成员及其子节点。通过深度优先搜索遍历家族树，计算每个世代的人口，找出最大人口的世代并输出其人口数和世代级别。示例输入和输出展示了算法如何工作。

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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

#include<iostream>
#include<vector>
using namespace std;
vector<int> id[100];
int book[100];
void dfs(int N, int depth) {
	book[depth]++;
	for (int i = 0; i < id[N].size(); i++)
		dfs(id[N][i], depth + 1);
}
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int ID, K;
		cin >> ID >> K;
		for (int j = 0; j < K; j++) {
			int temp;
			cin >> temp;
			id[ID].push_back(temp);
		}
	}
	dfs(1, 0);
	int maxnum = 0, maxdepth = 0;
	for (int k = 0; k < 100; k++) {
		if (book[k] > maxnum) {
			maxnum = book[k];
			maxdepth = k;
		}
	}
	cout << maxnum << " " << maxdepth + 1;
	return 0;
}