PAT1094

最新推荐文章于 2024-12-18 01:39:59 发布

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本文介绍了一种算法，用于解决家谱树中寻找人口最多的代际问题。输入包含家庭成员总数及子嗣信息，输出该代的最大人口数及其层级。

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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 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


#include
#include
#include
#include
using namespace std;
#define MAX 10000
vector >v(100);
int level,ans_node,visit[MAX],max_level=0; 
int search(){
	queueq;
	level=0;
	int level_node=1;
	ans_node=1;
	int node_number=0;
	visit[1]=1;
	q.push(level_node);
	while(!q.empty()){
		int node=q.front();
		q.pop();
		node_number++; 
		for(int i=0;i=ans_node){
				ans_node=node_number;
				max_level=level;
			}
			node_number=0;
		}
	}
	return 1;
}
int main(){
	int node_number,child_number;
	cin>>node_number>>child_number;
	for(int i=1;i<=child_number;i++){
		int parent,count;
		cin>>parent>>count;
		for(int i=1;i<=count;i++){
			int node;
			cin>>node;
			v[parent].push_back(node);
			v[node].push_back(parent);
		}
	}
	memset(visit,0,sizeof(visit));
	ans_node=1;
	max_level=0;
	search();
	cout<