PAT Nowcoder Advanced 1005 The Largest Generation
题目
The Largest Generation (25)
题目描述
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.
输入描述:
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.
输出描述:
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.
输入例子:
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
输出例子:
9 4
理解与算法
这道题跟PAT模拟题的1004基本一致,不过从求叶子结点的数量变成了求节点最多的层的深度,以及节点的数量。
代码与实现
//
// Created by tanknee on 2021/1/27.
//
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> nodes[100]; // 每个元素代表一个节点链表
int pedigree[100], population[101]; // 族谱树中每一层的叶子结点的数量
int pedigree_depth = -1; // 族谱树的最大深度
/**
* @param index 下标
* @param depth 深度
*/
void dfs(int index, int depth) {
if (!nodes[index].empty()) {
population[depth] += nodes[index].size();
}
// 遍历该节点的所有子节点
for (int i : nodes[index]) {
// 因为往下走了一层,所以深度加1
dfs(i, depth + 1);
}
}
int main() {
int N, M, node, num, child, max = 0, max_i = 0;
// 处理第一行
cin >> N >> M;
// 遍历所有的非叶节点,构建节点链表
for (int i = 0; i < M; ++i) {
cin >> node >> num;
for (int j = 0; j < num; ++j) {
cin >> child;
nodes[node].push_back(child);
}
}
// 对族谱树进行深度优先遍历
dfs(1, 0);
for (int i = 0; i < N; ++i) {
if (population[i] > max) {
max = population[i];
max_i = i + 2;
}
}
cout << max << " " << max_i;
return 0;
}