1094 The Largest Generation (25 分)（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

生词

英文	解释
hierarchy	等级制度
pedigree tree	谱系树

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～

原文链接：https://blog.csdn.net/liuchuo/article/details/52213503

题解

#include <bits/stdc++.h>

using namespace std;
const int maxn=110;
vector<int> node[maxn];
int hashTable[maxn];//树的静态写法，node[i]存放结点i的孩子结点编号

void DFS(int index,int level)
{
    hashTable[level]++;//第level层的节点个数加1
    for(int j=0;j<node[index].size();j++){
        DFS(node[index][j],level+1);
    }
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,m,parent,k,child;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>parent>>k;
        for(int j=0;j<k;j++){
            cin>>child;
            node[parent].push_back(child);
        }
    }
    DFS(1,1);
    int maxLevel=-1,maxValue=0;
    for(int i=1;i<maxn;i++){
        if(hashTable[i]>maxValue){
            maxValue=hashTable[i];
            maxLevel=i;
        }
    }
    cout<<maxValue<<" "<<maxLevel<<endl;
    return 0;
}