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; }
dfs
FH