1094. The Largest Generation (25)

最新推荐文章于 2024-07-16 16:03:51 发布

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原文链接：http://www.cnblogs.com/grglym/p/8001097.html

本文介绍了一种通过构建家族成员的树形结构来确定拥有最多成员的家族代数的方法。输入包括家族成员总数及各成员的子代信息，输出则是人口最多的代数及其编号。

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<iostream>
#include<vector>
#include<queue>
#include<cstdio>
using namespace std;
int main(){
	vector<int>vt[102];
	int n,m;
	scanf("%d%d",&n,&m);
	int i,j;
	int id,k;
	int val;
	for(i=0;i<m;i++){
		scanf("%d%d",&id,&k);
		for(j=0;j<k;j++){
			scanf("%d",&val);
			vt[id].push_back(val);
		}
	}
	queue<int>q;
	q.push(1);
	int level=1;
	int lev=1;
	int largestnum=1;
	int cnt=0;
	int last=1,start=0;
	while(!q.empty()){
		int tmp=q.front();
		q.pop();
		start++;
		int size=vt[tmp].size();
		for(i=0;i<size;i++){
			q.push(vt[tmp][i]);
			cnt++;
		}
		if(start==last){
			lev++;
			start=0;
			if(cnt>largestnum){
				largestnum=cnt;
				level=lev;
			}
			last=q.size();
			cnt=0;
			//cout<<cnt<<" "<<start<<" "<<last<<endl;
		}
	}
	printf("%d %d",largestnum,level);
	return 0; 
}

转载于:https://www.cnblogs.com/grglym/p/8001097.html