1094 The Largest Generation(25 分)
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
Analysis:
1. use vector to record the tree.
2. bfs and record the level of each node according to its parent
3. count the node of each level an find the largest
C++:
#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
vector<int>v[100];
queue<int>q;
int level[100];
int main()
{
//freopen("1094 The Largest Generation.txt","r",stdin);
int n,m;
cin>>n>>m;
for(int i=0;i<m;i++){
int num,k,child;
cin>>num>>k;
while(k--){
cin>>child;
v[num].push_back(child);
}
}
q.push(1);
level[1]=1;
while(!q.empty()){
int num=q.front();
int l=level[num];
q.pop();
for(int i=0;i<v[num].size();i++){
q.push(v[num][i]);
level[v[num][i]]=l+1;
}
}
int count1[100]={0},maxl=0,j;
for(int i=1;i<=n;i++)
count1[level[i]]++;
for(int i=1;i<100;i++){
if(maxl<count1[i]){
maxl=count1[i];
j=i;
}
}
cout<<maxl<<" "<<j;
return 0;
}