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
常规题吧,PAT很多题都包括层序遍历了
我还是习惯用vector<vector> vec(N)来保存边,然后使用队列遍历每一层
代码:
#include <iostream>
#include <algorithm>
using namespace std;
#include <unordered_map>
#include <queue>
#include <vector>
int main()
{
//int sn[100][100]={0};
vector<vector<int>> sn(100);
int N,M;
cin>>N>>M;
int sign[N]={0};
for(int i=0;i<M;i++){
int id;
cin>>id;
int num;
cin>>num;
for(int m=0;m<num;m++){
int get;
cin>>get;
sn[id].push_back(get);
sign[get]=-1;
}
}
int root;
for(int i=1;i<=N;i++){
if(sign[i]==0){
root=i;
break;
}
}
int level=1;
int maxx=1;
int maxLevel=1;
queue<int> que;
que.push(root);
while(!que.empty()){
queue<int> get;
while(!que.empty()){
int getNum=que.front();
for(int i=0;i<sn[getNum].size();i++){
get.push(sn[getNum][i]);
}
que.pop();
}
level++;
if(get.size()>maxx){
maxx=get.size();
maxLevel=level;
}
que=get;
}
cout<<maxx<<' '<<maxLevel;
return 0;
}