PAT-A 1094. The Largest Generation

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

这道题其实考的是图的BFS遍历，如果当成树来看的话会比较麻烦，因为孩子的个数是不确定的。还是看成一个特殊的图来看待比较方便。C++当中有很多很好用的容器，比如queue、stack等等，可怜我之前用C，全部都要自己写，真是蠢到姥姥家了。这道题比较麻烦的是计算每一层的结点个数，用两重for循环可以解决这个问题。

程序代码：

#include<iostream>
using namespace std;
#define MAX 100
#include<queue>
#include<vector>
vector <int> num[MAX];
bool visited[MAX]={0};
void BFS(int s);
int sum = 0;
int max_level =1;
int main()
{
	int N,M; //N总人数，M有孩子的人数
	cin>>N>>M;
	int ID,k;
	int tmp;
	for(int i=0;i<M;i++)
	{
		cin>>ID>>k;
		for(int j=0;j<k;j++)
		{
			cin>>tmp;
			num[ID].push_back(tmp);				
		}
	}
	BFS(1);
	cout<<sum<<" "<<max_level;
	return 0;
}
void BFS(int s)
{
	int level =1;
	//max_level = 1;
	int cnt,cnt1=1;
	//max = 0;
	queue<int> q;
	q.push(s);
	visited[s]=true;
	int tmp;
	while(!q.empty())
	{
		cnt=cnt1;
		cnt1=0;
		if(cnt>sum)
		{
			sum = cnt;
			max_level = level;
		}
		for(int j=0;j<cnt;j++)
		{
			tmp = q.front();
			q.pop();
			visited[tmp]= true;
			for(int i=0;i<num[tmp].size();i++)
			{
				q.push(num[tmp][i]);
				cnt1++;
			}
		}
		level++;
	}
}