PAT-A 真题 – 1094 The Largest Generation

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：给出一颗家庭关系树，01为根，算出人最多的一代，并求出由多少人

这是一个树的层序遍历问题，使用带层级的BFS算法即可。

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#define MAXN 101
using namespace std;
typedef struct{
  int id, layer;
} node;
vector<int> T[MAXN];
int fuck[MAXN] = {0};

void BFS(int root){
  queue<node> Q;
  Q.push({root, 1});
  while(!Q.empty()){
    node it = Q.front();
    Q.pop();
    fuck[it.layer]++;
    for(int i = 0; i < T[it.id].size(); i++){
      node new_fuck;
      new_fuck.id = T[it.id][i];
      new_fuck.layer = it.layer + 1;
      Q.push(new_fuck);
    }
  }
}

int main(){
  int N, M;  //n为节点数量，m为非叶子节点数量
  cin >> N >> M;
  for(int i = 0; i < M; i++){
    int root, sum;
    cin >> root >> sum;
    for(int j = 0; j < sum; j++){
      int _;
      cin >> _;
      T[root].push_back(_);
    }
  }
  BFS(1);
  int max = 0, layer = 0;
  for(int i = 0; i < MAXN; i++){
    if(fuck[i] > max){
      max = fuck[i];
      layer = i;
    }
  }
  cout << max << ' ' << layer << endl;
  return 0;
}