1094. The Largest Generation

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 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

存好dfs就可以了。

#include<iostream>
#include<vector>

using namespace std;

vector <int> ch[100];
int cnt[101];

void dfs(int num,int g)
{
	cnt[g]++;
	if(ch[num].empty())
		return;
	for(int i=0;i<ch[num].size();++i)
		dfs(ch[num][i],g+1);
}

int main()
{
	int n,m;
	cin>>n>>m;
	int id,chn,chid;
	for(int i=0;i<m;++i)
	{
		cin>>id>>chn;
		for(int j=0;j<chn;++j)
		{
			cin>>chid;
			ch[id].push_back(chid);
		}
	}
	for(int i=1;i<=n;++i)
		cnt[i]=0;
	dfs(1,1);
	int maxl=1;
	for(int i=1;i<=n;++i)
	{
		if(cnt[i]>cnt[maxl])
			maxl=i;
	}
	cout<<cnt[maxl]<<" "<<maxl;
	return 0;
}