1094 The Largest Generation (25point(s))
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
DFS
#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
int n,m,maxNum=0,maxDepth=1;
vector<int> Node[maxn];
int hashTable[maxn]={
0};
void DFS(int index,int depth){
hashTable[depth]++;
for(int i=0;i<Node[index].size();++i)
DFS(Node[index][i],depth+1);
}
int main(){
scanf("%d%d",&n,&m);
int father,child,k;
for(int i=1;i<=m;++i){
scanf("%d%d",&father,&k);
for(int i=0;i<k;++i){
scanf("%d",&child);
Node[father].push_back(child);
}
}
DFS(1,1);
for(int i=1;i<=maxn;++i){
if(hashTable[i]>maxNum){
maxNum=hashTable[i];
maxDepth=i;
}
}
printf("%d %d\n",maxNum,maxDepth);
}
BFS
#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
int n,m,maxNum=0,maxDepth=1;
struct node{
vector<int> child;
int depth;
}Node[maxn];
int hashTable[maxn]={
0};
void BFS(int index){
queue<int> q;
q.push(index);
Node[index].depth=1;
while(!q.empty()){
int now=q.front();
hashTable[Node[now].depth]++;
q.pop();
for(int i=0;i<Node[now].child.size();++i){
int child=Node[now].child[i];
Node[child].depth=Node[now].depth+1;
q.push(child);
}
}
}
int main(){
scanf("%d%d",&n,&m);
int father,child,k;
for(int i=1;i<=m;++i){
scanf("%d%d",&father,&k);
for(int i=0;i<k;++i){
scanf("%d",&child);
Node[father].child.push_back(child);
}
}
BFS(1);
for(int i=1;i<=maxn;++i){
if(hashTable[i]>maxNum){
maxNum=hashTable[i];
maxDepth=i;
}
}
printf("%d %d\n",maxNum,maxDepth);
}