PAT 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

简述一下题目，从第二行开始输入的数，第一个是父亲id，紧接着的k是孩子个数，后面k个是孩子的id

现在给你一个家族的信息，让你求出同一辈的人最多有多少个，并且算出这些人是第几代

很明显的并查集。用途很广

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int father[1000];
int n,m;
int find(int x){
if(x==father[x]) return x;
return father[x]=find(father[x]);
}
int main(){
cin>>n>>m;
int s=m;
for(int i=1;i<=n;i++)  father[i]=i;
while(s--){
int f,k;
cin>>f>>k;
for(int i=0;i<k;i++){
int num;
cin>>num;
father[num]=f;
}
}
int sum=0;
int len[1000+1]={0};
int max=0,num=0;
for(int i=1;i<=n;i++){
int b=i;
sum=0;
while(b!=father[b]){
b=father[b];
sum++;//sum纪录的是第几代孩子，然后直接存入len数组
}
len[sum]++;
}
for(int i=0;i<=m;i++){
if(max<len[i]){//找到同一辈最大的个数，纪录第几代
max=len[i];
num=i;
}
}
cout<<max<<" "<<num+1;//注意sum是从0开始的，要+1
return 0;
}

广搜

遍历即可

#include<iostream>
#include<queue>
#include<vector>
#include<stack>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<cstdio>
using namespace std;
typedef pair<int,int> P;
int main(){
int n,m;
cin>>n>>m;
vector<int> v[200];
for(int i=0;i<m;i++){
int num,len;
scanf("%d%d",&num,&len);
for(int j=0;j<len;j++){
int fz;
scanf("%d",&fz);
v[num].push_back(fz);
}
}
queue<P> que;
que.push({1,1});
int a[200]={0};
a[1]=1;
while(que.size()){
P p=que.front();
que.pop();
int num=p.first;
int len=p.second;
for(int i=0;i<v[num].size();i++){
int fz=v[num][i];
que.push({fz,len+1});
a[len+1]++;
}
}
int maxn=0,index=0;
for(int i=0;i<101;i++){
if(a[i]>maxn) maxn=a[i],index=i;
}
cout<<maxn<<" "<<index;
return 0;
}