• PAT (Advanced Level) Practice 1094 The Largest Generation (25分) (BFS改进)


    1.题目

    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

    2.题目分析

    思路是使用BFS,将在一层的节点统一放在一个vector中,计算个数并记录,之后进入下一层,当所有节点都访问过一遍后退出。

    3.代码

    #include<iostream>
    #include<vector>
    #include<queue>
    using namespace std;
    vector<int>list[110];
    int amount[110];//每层的节点个数
    int counts = 2;//amount的遍历变量,从2开始,第一层设为1(根节点)
    int total = 0;//遍历过的总节点数
    void DFS(int root,int n)
    {
    	queue<int>out;
    	out.push(root);
    	amount[1] = 1;//第一层1个节点
    	while (1)
    	{
    		vector<int>t;//先放在vector中,记录完个数后放在queue中
    		while (!out.empty())
    		{
    			int temp = out.front(); out.pop();
    			for (int i = 0; i < list[temp].size(); i++)
    			{
    				t.push_back(list[temp][i]);
    			}
    		}
    		amount[counts++] = t.size();
    		for (int i = 0; i < t.size(); i++)
    		{
    			out.push(t[i]); total++;
    		}
    		t.clear();
    		if (total == n - 1)break;//除了根节点剩下n-1个
    	}
    }
    int main()
    {
    	int n, m, k,a,b;
    	scanf("%d %d", &n, &m);
    	for (int i = 1; i <= m; i++)
    	{
    		scanf("%d", &a);
    		scanf("%d", &k);
    		for (int j = 1; j <= k; j++)
    		{
    			scanf("%d", &b);
    			list[a].push_back(b);
    		}
    	}
    	DFS(1, n);
    	int max = -1, maxi;
    	for (int i = 1; i <= n; i++)
    	{
    		if (max < amount[i]) { max = amount[i]; maxi = i; }
    	}
    	printf("%d %d", max, maxi);
    }
  • 相关阅读:
    Orleans介绍
    Orleans入门
    mongodb for windows安装
    ASP.NET Identity
    OWIN与Katana
    手动搭建ABP2.1.3 Zero——基础框架
    ABP学习笔记
    ABP-Zero模块
    ABP单元测试
    ABP-JavaScript API
  • 原文地址:https://www.cnblogs.com/Jason66661010/p/12788837.html
Copyright © 2020-2023  润新知