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 记录每个点的父亲，然后算每个点的高度然后记录每层人数，统计记录。代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>
using namespace std;
int nn,n,f[105],num[105];
int id,k,subid,m,t;
int height(int k)
{
if(k == 0)return 0;
return height(f[k]) + 1;
}
int main()
{
cin>>nn>>n;
for(int i = 0;i < n;i ++)
{
cin>>id>>k;
for(int j = 0;j < k;j ++)
{
cin>>subid;
f[subid] = id;
}
}
for(int i = 1;i <= nn;i ++)
{
int d = height(i);
num[d] ++;
if(num[d] > m)m = num[d],t = d;
}
cout<<m<<‘ ‘<<t;
}
原文地址：https://www.cnblogs.com/8023spz/p/8395166.html