【PAT】【Advanced Level】1094. The Largest Generation (25)

最新推荐文章于 2019-03-27 23:19:34 发布

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最新推荐文章于 2019-03-27 23:19:34 发布

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本文提供PAT A 1094题的解析与代码实现，通过构建家族成员的树形结构，利用BFS遍历算法找出人口最多的代数。

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1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1094

思路：

vector存储子节点，然后BFS/DFS均可。

注意边界

CODE：

#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#define N 110
using namespace std;
vector<int> ch[N];
queue<int> qu;
int bf[N];
int f[N];
int main()
{
	memset(f,0,sizeof(f));
	memset(bf,0,sizeof(bf));
	int n,m;
	cin>>n>>m;
	for (int i=0;i<m;i++)
	{
		int num,k;
		cin>>num>>k;
		for (int j=0;j<k;j++)
		{
			int chi;
			cin>>chi;
			ch[num].push_back(chi);
		}
	}
	bf[1]=1;
	qu.push(1);
	//f[1]=1;
	while (!qu.empty())
	{
		int now=qu.front();
		qu.pop();
		for (int i=0;i<ch[now].size();i++)
		{
			bf[ch[now][i]]=bf[now]+1;
			qu.push(ch[now][i]);
		}
	}
	int mx=-1;
	for (int i=1;i<=n;i++)
	{
		mx=max(mx,bf[i]);
		f[bf[i]]++;
	}
	int maxn=-1;
	int maxi=0;
	for (int i=1;i<=mx;i++)
	{
		if (f[i]>maxn)
		{
			maxi=i;
			maxn=f[i];
		}
	}
	cout<<maxn<<" "<<maxi;
	return 0;
}