PAT甲级1094
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<vector>
using namespace std;
#define maxn 110
vector<int>node[maxn];
int n, m, deep[maxn] = { 0 }, maxh = 0;
void DFS(int index, int h)
{
deep[h]++;
if(node[index].size()==0)
{
if(maxh<h)
maxh=h;
return;
}
for (int i = 0; i < node[index].size(); i++)
DFS(node[index][i], h + 1);
}
int main()
{
cin >> n >> m;
int par, k, child;
for (int i = 0; i < m; i++)
{
cin >> par >> k;
for (int j = 0; j < k; j++)
{
cin >> child;
node[par].push_back(child);
}
}
DFS(1, 1);
int maxhnum = 0, pos;
for (int i = 1; i <=maxh; i++)
{
if (deep[i] > maxhnum)
{
pos = i;
maxhnum = deep[i];
}
}
cout << deep[pos] << " " << pos << endl;
return 0;
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