PAT甲级——1094 The Largest Generation (树的遍历)

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博客围绕族谱树问题展开，给定家族成员信息，要求找出人数最多的一代。输入包含家族成员总数和有孩子的成员数及相关信息，输出为该代人数和层级。解题思路是用层序遍历标记层数，通过数组映射层数和人数关系来求解。

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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

题目大意：一个家族里面的人的ID编号是从01到N，每个人有若干个孩子，形成一棵族谱树，同一层的人是同辈关系，找出人数最多的那一层并且输出该层的人数。

思路：层序遍历（BFS思想）标记每个人所在的层数，另开一个数组 ans[level] = num 用于映射层数和人数的关系，树遍历完成后再遍历ans数组就能找到人数最多的那一层。

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

struct node {
	int level = 0;
	vector<int> child;
};
vector<node> tree;
vector<int> ans;
void getLevel();
int main()
{
	int N, M, ID, K;
	scanf("%d%d", &N, &M);
	tree.resize(N + 1);
	ans.resize(N + 1, 0);
	for (int i = 0; i < M; i++) {
		scanf("%d%d", &ID, &K);
		tree[ID].child.resize(K);
		for (int j = 0; j < K; j++)
			scanf("%d", &tree[ID].child[j]);
	}
	getLevel();
	int level = 0, num = 0;
	for (int i = 0; i < ans.size(); i++) {
		if (num < ans[i]) {
			num = ans[i];
			level = i;
		}
	}
	printf("%d %d\n", num, level);
	return 0;
}

void getLevel() {
	int ID = 1;
	queue<int> Q;
	tree[ID].level = 1;
	Q.push(ID);
	while (!Q.empty()) {
		ID = Q.front();
		ans[tree[ID].level]++;
		Q.pop();
		int childID;
		for (int i = 0; i < tree[ID].child.size(); i++) {
			childID = tree[ID].child[i];
			tree[childID].level = tree[ID].level + 1;
			Q.push(childID);
		}
	}
}