1094. The Largest Generation (25)
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
1 #include<string> 2 #include<iostream> 3 #include<map> 4 #include<stdio.h> 5 #include<vector> 6 using namespace std; 7 8 struct node 9 { 10 vector<int> child; 11 }; 12 13 node Tree[110]; 14 int Level[110]; 15 int MAX = -1; 16 void DFS(int n,int level) 17 { 18 if(MAX < level) 19 MAX = level; 20 ++Level[level]; 21 for(int i = 0 ;i < Tree[n].child.size() ;++i) 22 DFS(Tree[n].child[i],level+1); 23 } 24 int main() 25 { 26 int n,m,tid,num,id; 27 scanf("%d%d",&n,&m); 28 for(int i = 0 ;i < m ;++i) 29 { 30 scanf("%d%d",&id,&num); 31 for(int k = 0;k < num;++k) 32 { 33 scanf("%d",&tid); 34 Tree[id].child.push_back(tid); 35 } 36 } 37 DFS(1,1); 38 int maxL = -1; 39 int index = -1; 40 for(int i = 1; i <= MAX;++i) 41 { 42 if(Level[i] > maxL) 43 { 44 maxL = Level[i]; 45 index = i; 46 } 47 } 48 printf("%d %d\n",maxL,index); 49 return 0; 50 }