PAT (Advanced Level) Practice 1094 The Largest Generation (25分) （BFS改进）

1.题目

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

2.题目分析

思路是使用BFS，将在一层的节点统一放在一个vector中，计算个数并记录，之后进入下一层，当所有节点都访问过一遍后退出。

3.代码

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<int>list[110];
int amount[110];//每层的节点个数
int counts = 2;//amount的遍历变量，从2开始，第一层设为1（根节点）
int total = 0;//遍历过的总节点数
void DFS(int root,int n)
{
	queue<int>out;
	out.push(root);
	amount[1] = 1;//第一层1个节点
	while (1)
	{
		vector<int>t;//先放在vector中，记录完个数后放在queue中
		while (!out.empty())
		{
			int temp = out.front(); out.pop();
			for (int i = 0; i < list[temp].size(); i++)
			{
				t.push_back(list[temp][i]);
			}
		}
		amount[counts++] = t.size();
		for (int i = 0; i < t.size(); i++)
		{
			out.push(t[i]); total++;
		}
		t.clear();
		if (total == n - 1)break;//除了根节点剩下n-1个
	}
}
int main()
{
	int n, m, k,a,b;
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= m; i++)
	{
		scanf("%d", &a);
		scanf("%d", &k);
		for (int j = 1; j <= k; j++)
		{
			scanf("%d", &b);
			list[a].push_back(b);
		}
	}
	DFS(1, n);
	int max = -1, maxi;
	for (int i = 1; i <= n; i++)
	{
		if (max < amount[i]) { max = amount[i]; maxi = i; }
	}
	printf("%d %d", max, maxi);
}