挽救自信心的题...

#include <cstdio>
#include <vector>
#include <map>
const int MAXN = 100;int N, M, K, id, maxP, maxLevel;
std::vector<int> vec[MAXN];
std::map<int, int> mp;void dfs(int p, int l){for(int i = 0; i < vec[p].size(); ++i){mp[l + 1]++;dfs(vec[p][i], l + 1);}
}int main(){scanf("%d %d", &N, &M);for(int i = 0; i < M; ++i){scanf("%d %d", &id, &K);vec[id].resize(K);for(int j = 0; j < K; ++j){scanf("%d", &vec[id][j]);}}mp[1]++;dfs(1, 1);maxP = 0;for(auto it = mp.begin(); it != mp.end(); ++it){if(it->second > maxP){maxP = it->second;maxLevel = it->first;}}printf("%d %d", maxP, maxLevel);return 0;
}

题目如下：

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