The Largest Generation (25)

最新推荐文章于 2024-06-02 16:33:44 发布

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最新推荐文章于 2024-06-02 16:33:44 发布

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本文链接：https://blog.youkuaiyun.com/u012816621/article/details/50669969

本文介绍了一个算法问题，即如何通过给定的家庭成员信息找到人口最多的那一辈及其对应的辈分。通过对家庭成员之间的父子关系建立映射，并采用递归的方式计算各代的人口数量，最终确定了人口最多的代。

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The Largest Generation (25)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
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.

输入描述:
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.
输出描述:
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.
输入例子:
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

输出例子:
9 4

#include <iostream>
#include <malloc.h>
#include <fstream>
using namespace std;

/*
    题目大意：给出家族的所有人数目和有孩子的人的数目，并且把每个人进行编号，找出人最多的那一代一共多
    少人并且输出该代是第几代。
    构建数组family[child] = parent; 并自低向顶计算代数。
*/
int getLayer(int *family, int child)
{
    int gen = 1;
    //依次往上取值
    while (family[child] != child)
    {
        child = family[child];
        gen++;
    }
    return gen;
}

int main()
{
    int n,m;
    //ifstream fin("test.txt");
    cin >> n >> m;
    //孩子为下标，父母为值
    int *family = (int *)malloc((n+1)*sizeof(int));
    family[1] = 1;  //赋值根节点
    for (int i = 0; i < m; i++)
    {
        int parent, childNum;
        cin >> parent >> childNum;
        for (int j = 0; j < childNum; j++)
        {
            int child;
            cin >> child;
            family[child] = parent;
        }
    }
    //计算某代多少人
    int *layer = (int *)malloc((n + 1)*sizeof(int));
    for (int i = 1; i <= n; i++)
        layer[i] = 0;
    for (int i = 1; i <= n; i++)
        layer[getLayer(family, i)]++;   //把得到的相应层次人数累加；
    int maxPeople=1,p=1;
    for (int i = 1; i <= n; i++)
    {
        if (layer[i] > maxPeople)
        { 
            maxPeople = layer[i];
            p = i;
        }
    }
    cout << maxPeople << ' ' << p;
    //getchar();
}