PAT 1094. The Largest Generation (层级遍历)
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
分析
用邻接表建树,层级遍历更新数据,用last,tail更新层级
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
int main(){
int n,m,father,son,num,last=1,tail,cnt=0,max=-1,level,tag=1;
cin>>n>>m;
vector<int> family[n+1];
for(int i=0;i<m;i++){
cin>>father>>num;
for(int j=0;j<num;j++){
cin>>son;
family[father].push_back(son);
}
}
queue<int> q;
q.push(1);
while(!q.empty()){
int t=q.front();
q.pop();
cnt++;
for(int i=0;i<family[t].size();i++){
q.push(family[t][i]);
tail=family[t][i];
}
if(t==last){
last=tail;
if(cnt>max){
max=cnt;
level=tag;
}
tag++;
cnt=0;
}
}
cout<<max<<" "<<level;
return 0;
}