PAT 甲级 1094 The Largest Generation

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https://pintia.cn/problem-sets/994805342720868352/problems/994805372601090048

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

代码：

#include <bits/stdc++.h>
using namespace std;int N, M;
int vis[110], num[110];
vector<int> v[110];
int depth = -1;void dfs(int root, int step) {
if(v[root].size() == 0) {
depth = max(depth, step);
} vis[root] = 1;
num[step] ++;
for(int i = 0; i < v[root].size(); i ++) {
if(vis[v[root][i]] == 0) {
dfs(v[root][i], step + 1);
}
}
}int main() {
scanf("%d%d", &N, &M);
memset(vis, 0, sizeof(vis));
for(int i = 0; i < M; i ++) {
int id, K, x;
scanf("%d%d", &id, &K);
for(int k = 0; k < K; k ++) {
scanf("%d", &x);
v[id].push_back(x);
}
} dfs(1, 1);
int temp, ans = INT_MIN;
for(int i = 0; i <= depth; i ++) {
if(num[i] > ans) {
ans = num[i];
temp = i;
}
}
/*for(int i = 1; i <= depth; i ++)
printf("%d ", num[i]);*/
printf("%d %d\n", ans, temp);
//printf("%d\n", depth);
return 0;
}