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=;
vector<int> fa[maxn];
int main(){
int n,m;
scanf("%d %d",&n,&m);
for(int i=;i<=m;i++){
int root,k;
scanf("%d %d",&root,&k);
for(int j=;j<k;j++){
int ch;
scanf("%d",&ch);
fa[root].push_back(ch);
}
}
queue<int> q;
q.push();
int maxm=,lvl=,max_l=;
while(!q.empty()){
queue<int> child;
int num=;
while(!q.empty()){
int now = q.front();
q.pop();
for(int i=;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