Pat(Advanced Level)Practice--1094(The Largest Generation)

Pat1094代码

题目描述：

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

AC代码：

#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>#include<queue>#include<vector>#define MAXN 105using namespace std;vector<int> v[MAXN];queue<int> q;int visited[MAXN];int main(int argc,char *argv[]){int n,m;scanf("%d %d",&n,&m);for(int i=0;i<m;i++){int f,k,c;scanf("%d %d",&f,&k);for(int j=0;j<k;j++){scanf("%d",&c);v[f].push_back(c);}}int max=1,level=1;int count=0,cur_level=0;q.push(1);int endNode=q.back();while(!q.empty()){int index=q.front();q.pop();for(int i=0;i<v[index].size();i++){int j=v[index][i];if(!visited[j]){visited[j]=1;q.push(j);}}if(index==endNode){cur_level++;count=q.size();if(count>max){max=count;level=cur_level+1;}endNode=q.back();}}printf("%d %d\n",max,level);return 0;}