PAT A1094 The Largest Generation （25 分）——树的bfs遍历

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 IDto 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 <stdio.h>
#include <vector>
#include <queue>
using namespace std;
const int maxn=110;
vector<int> fa[maxn];
int main(){
  int n,m;
  scanf("%d %d",&n,&m);
  for(int i=1;i<=m;i++){
    int root,k;
    scanf("%d %d",&root,&k);
    for(int j=0;j<k;j++){
      int ch;
      scanf("%d",&ch);
      fa[root].push_back(ch);
    }
  }
  queue<int> q;
  q.push(1);
  int maxm=1,lvl=1,max_l=1;
  while(!q.empty()){
    queue<int> child;
    int num=0;
    while(!q.empty()){
      int now = q.front();
      q.pop();
      for(int i=0;i<fa[now].size();i++){
        child.push(fa[now][i]);
        num++;
      }
    }
    lvl++;
    if(num>maxm){
      maxm=num;
      max_l=lvl;
    }
    while(!child.empty()){
      q.push(child.front());
      child.pop();
    }
  }
  printf("%d %d",maxm,max_l);
}

注意点：统计每层个数，用两个队列实现，同时统计个数和层数，一层全遍历完，再把下一层加入到队列中去