PAT甲级——A1094 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 (<) 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 (<) 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 (>) 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

#include <iostream>
#include <queue>
#include <vector>
using namespace std;
int N, M, maxN = 1, resL = 1, root = 1, level[105] = { 0 }, manN[105] = { 0 };
vector<int>man[105];
void BFS()
{
    queue<int>q;
    q.push(root);
    level[root] = 1;
    manN[level[root]]++;
    while (!q.empty())
    {
        root = q.front();
        q.pop();
        int temp = 0;
        for (auto v : man[root])
        {
            level[v] = level[root] + 1;
            manN[level[v]]++;//记录每一层的人数
            if (man[v].size() > 0)
                q.push(v);
        }
    }    
}

void DFS(int s,int l)
{
    manN[l]++;//l层的人数
    for (auto v : man[s])
        DFS(v, l + 1);
}

int main()
{
    cin >> N >> M;
    for (int i = 0; i < M; ++i)
    {
        int a, b, k;
        cin >> a >> k;
        for (int j = 0; j < k; ++j)
        {
            cin >> b;
            man[a].push_back(b);
        }
    }
    //BFS();
    DFS(1, 1);
    for (int i = 1; i <= N; ++i)
    {
        if (maxN < manN[i])
        {
            maxN = manN[i];
            resL = i;
        }
    }    
    cout << maxN << " " << resL << endl;
    return 0;
}