PAT A1094 The Largest Generation(25)【最大的一代，DFS，BFS】

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

题意

统计树节点最多的层数和节点数

思路1

• 使用 vector[N] 邻接表存储树

• DFS 遍历到每一层将对应的计数器加1

代码1

#include <iostream>
#include <vector>
using namespace std;
const int N = 110;

vector<int> v[N]; // 孩子节点
int cnt[N];  // 每一层数量

void dfs(int u, int depth)
{
    cnt[depth]++;

    for(int i = 0; i < v[u].size(); i++)
    {
        dfs(v[u][i], depth + 1);
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++)
    {
        int id, k;
        scanf("%d%d", &id, &k);
        for(int j = 0; j < k; j++)
        {
            int son;
            scanf("%d", &son);
            v[id].push_back(son);
        }
    }

    dfs(1, 1); // 根结点序号为1，第一层

    int max_v = -1, max_idx = -1;
    for(int i = 1; i < N; i++)
    {
        if(cnt[i] > max_v)
        {
            max_v = cnt[i];
            max_idx = i;
        }
    }
    printf("%d %d\n", max_v, max_idx);
    return 0;
}

思路2

代码2