PAT甲级1094

Posted 2022-08-30 wx630c98f24f6b8

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 IDs 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

#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 100;
vector<int> children[maxn];
int level[maxn] =  0 ;
int N, M;
void BFS(int root)

  queue<int> Q;
  Q.push(root);
  level[root] = 1;
  while (!Q.empty())
  
    int t = Q.front();
    Q.pop();
    if (children[t].size())
    
      for (int i = 0; i < children[t].size(); i++)
      
        Q.push(children[t][i]);
        level[children[t][i]] = level[t] + 1;
      
    
  

int hashtable[maxn] =  0 ;
int main()

  scanf("%d%d", &N,&M);
  int father,k, child;
  for (int i = 0; i < M; i++)
  
    scanf("%d %d", &father,&k);
    for (int j = 0; j < k; j++)
    
      scanf("%d", &child);
      children[father].push_back(child);
    
  
  BFS(1);
  for (int i = 1; i <=N; i++)
  
    hashtable[level[i]]++;
  
  int maxL = -1, maxl;
  for (int i = 1; i <= N; i++)
  
    if (maxL < hashtable[i])
    
      maxL = hashtable[i];
      maxl = i;
    
  
  printf("%d %d", maxL, maxl);
  return 0;