PAT (Advanced Level) 1094 The Largest Generation

最新推荐文章于 2021-09-14 00:05:33 发布

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最新推荐文章于 2021-09-14 00:05:33 发布

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本文介绍了一种算法，用于解决在给定的家族树结构中找出人口最多的世代的问题。通过深度优先搜索（DFS）遍历每个节点，并计算各世代的人口数量，最终确定人口最多的世代及其人口数。

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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

Code:

#include <iostream>
#include <vector>
#include <algorithm>
#pragma warning(disable:4996)
using namespace std;

int n, m;
vector<vector<int>> vec;
vector<int> level_cnt;

void dfs(int index, int depth)
{
	level_cnt[depth]++;
	for (int i = 0; i < vec[index].size(); i++)
		dfs(vec[index][i], depth + 1);
}

int main()
{
	scanf("%d%d", &n, &m);
	vec.resize(n + 1); level_cnt.resize(n + 1, 0);
	for (int i = 0; i < m; i++)
	{
		int p, cn, c;
		scanf("%d%d", &p, &cn);
		for (int j = 0; j < cn; j++)
		{
			scanf("%d", &c);
			vec[p].push_back(c);
		}
	}
	dfs(1, 1);
	int g = 1, sum = 0;
	for (int i = 1; i < level_cnt.size() && level_cnt[i] > 0; i++)
	{
		if (level_cnt[i] > sum)
		{
			sum = level_cnt[i];
			g = i;
		}
	}
	printf("%d %d\n", sum, g);
	return 0;
}