PAT(甲级)1094

1094. The Largest Generation (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 1321 1 2301 4 03 02 04 0503 3 06 07 0806 2 12 1313 1 2108 2 15 1602 2 09 1011 2 19 2017 1 2205 1 1107 1 1409 1 1710 1 18

Sample Output:

9 4

#include <iostream>#include <deque>using namespace std;struct Person{Person(){num = 0;Next =NULL;}Person(int n){num = n;Next = NULL;}void Add(int n);int num;Person *Next;};void Person::Add(int n){Person *p = this;while(p->Next != NULL)    p = p->Next;p->Next= new Person(n);}Person persons[101];void MyBFS(deque<int> &q,int N){int count,count1,max;int gen=1;int genv=0;count = max =count1=0;q.push_back(1);q.push_back(-1);while(!q.empty()){int curr=q.front();if(curr != -1){count++;count1++;Person *p = persons[curr].Next;while(p != NULL){q.push_back(p->num);p =p->Next;}}else{genv++;q.push_back(-1);if(count > max){max = count;gen = genv;}count = 0;if(count1 ==N)        break;}q.pop_front();}cout <<max <<' ' <<gen <<endl;return;}int main(){int N,M;cin >>N >>M;int i,j,K,tmp,index;for(i=0;i<M;i++){cin >>index >>K;persons[index].num =K;for(j=0;j<K;j++){cin >>tmp;persons[index].Add(tmp);}}deque<int> q;MyBFS(q,N);return 0;}