PAT A1094. The Largest Generation (25) [BFS， DFS，树的遍历]

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
题⽬⼤意：输⼊树的结点个数N，结点编号为1~N，⾮叶⼦结点个数M，然后输出M个⾮叶⼦结点格⼦
的孩⼦结点的编号，求结点个数最多的⼀层，根结点的层号为1，输出该层的结点个数以及层号
分析：⽤DFS或者BFS，⽤DFS就⽤参数index和level标记当前遍历的结点的编号和层数，⼀个数组book
标记当前层数level所含结点数，最后遍历⼀遍数组找出最⼤值。注意 ： book[level]++;这句话是发⽣在
return语句判断之前的外⾯，即每遇到⼀个结点都要进⾏处理，⽽不是放在return语句的条件判断⾥⾯
～～
如果是BFS，就⽤⼀个数组level[i]标记i结点所处的层数，它等于它的⽗亲结点的level的值+1，⽤⼀个
数组book， book[i]标记i层所拥有的结点数，在遍历的时候每弹出⼀个结点就将当前结点的层数所对应
的book值+1，最后遍历⼀遍book数组找出最⼤拥有的结点数和层数

#include <cstdio>
#include <vector>
using namespace std;
vector<int> v[100];
int book[100];
void dfs(int index, int level) {
	book[level]++;
	for(int i = 0; i < v[index].size(); i++)
		dfs(v[index][i], level+1);
}
int main() {
	int n, m, a, k, c;
	scanf("%d %d", &n, &m);
	for(int i = 0; i < m; i++) {
		scanf("%d %d",&a, &k);
		for(int j = 0; j < k; j++) {
			scanf("%d", &c);
			v[a].push_back(c);
		}
	}
	dfs(1, 1);
	int maxnum = 0, maxlevel = 1;
	for(int i = 1; i < 100; i++) {
		if(book[i] > maxnum) {
			maxnum = book[i];
			maxlevel = i;
		}
	}
	printf("%d %d", maxnum, maxlevel);
	return 0;
}