1094 The Largest Generation (25 分)（深搜详解）

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

结尾无空行
Sample Output:

9 4

结尾无空行

一开始读题，以为是家庭成员，求有多少个家庭问题，上来就直接并查集了，结果写完才发现，此题要求的是，第几代人最多人数，然后求出来，画出树形图：

图画的有点草率了，见谅，从图中可以看出，第四代人的个数最多，一共有 9 个人，就是我们要输出的结果，所以，此题我们采用深度优先搜索进行解答。

定义数组 v 记录路径信息
定义 vis 判断是否访问过
定义 map<int, vector> 键指的是层数， 值指的是每层都有什么人。

首先，根据 m 求出所有的 v 图，需要注意的是，我们不确定谁才是根节点，所以，我这里开始用 vis 判断根节点，如果某个人没在孩子里出现，那么他就是祖先。

找到祖先，深搜，每次让那层人加入到 map 中。

最后一次遍历，求出 层数和最大人数即可。

#include<iostream>
#include<string>
#include<algorithm>
#include<bits/stdc++.h>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<deque>
#include<cctype>
#include<unordered_set>
#include<unordered_map>
#include<fstream>
#include<cstring>
using namespace std;
const int N = 101;
vector<int> v[N];
bool vis[N] = {false};
map<int, vector<int>> mp;
void Dfs(int x, int level) {
	mp[level].push_back(x);// x 是该层的人，加入 map
	vis[x] = 1;// 标记访问
	for (int i = 0; i < v[x].size(); i++) {
		if (!vis[v[x][i]]) {// 如果没访问过，深搜
			Dfs(v[x][i], level + 1);
		}
	}
}
int main () {
	int n, m;
	cin >> n >> m;
	while (m--) {
		int id, k;
		cin >> id >> k;
		for (int i = 0; i < k; i++) {
			int x;
			cin >> x;
			v[id].push_back(x);
			vis[x] = true;// 孩子标记
		}
	}
	for (int i = 1; i <= n; i++) {
		if (!vis[i]) {// 找到祖先
			memset(vis, 0, sizeof(vis));
			Dfs(i, 1);
			break;
		}
	}
	int sum = 0, level = 0;
	for (auto it = mp.begin(); it != mp.end(); it++) {
		if (it->second.size() > sum) {// 找到最大层数的人
			sum = it->second.size();
			level = it->first;
		}
	}
	cout << sum << " " << level;
	return 0;
}

标签：1094,25,vis,Generation,generation,level,int,include,ID
来源： https://blog.csdn.net/weixin_45962741/article/details/121487057