C++复习笔记--STL练习案例1（评委打分和员工分组）

目录

1--评委打分

1-1--案例描述

1-2--代码实现

2--员工分组

2-1--案例分析

2-2--代码实现

1--评委打分

1-1--案例描述

        有 A, B, C, D, E 共 5 名选手，由 10 个评委分别对每一名选手打分，去除评委的最高分和最低分，取剩下 8 名评委的平均分作为最终选手的评分；

1-2--代码实现

#include <iostream>
#include <queue>
#include <vector>
#include <string>
#include <algorithm>
#include <ctime>

class Person{
public:
    Person(std::string name, int score){
        this -> name = name;
        this -> score = score;
    }
    std::string name;
    int score;
};

void createPerson(std::vector<Person>& v){
    std::string nameSeed = "ABCDE";
    for(int i = 0; i < 5; i++){
        std::string name = "player";
        name += nameSeed[i];
        int score = 0;
        Person p(name, score);
        v.push_back(p);
    }
}

void setScore(std::vector<Person>&v){
    for(std::vector<Person>::iterator it = v.begin(); it != v.end(); it++){
        // 将评委的分数放入到deque容器中
        std::deque<int>d;
        for(int i = 0; i < 10; i++){
            int score = rand()%41 + 60; // 60 -100
            d.push_back(score);
        }
        std::cout<< "player: " << (*it).name << ", score: " << std::endl;
        for(std::deque<int>::iterator it = d.begin(); it != d.end(); it++){
            std::cout << *it << " ";
        }
        std::cout << std::endl;
        
        // 排序
        std::sort(d.begin(), d.end());
        // 去除最高分和最低分
        d.pop_back();
        d.pop_front();
        // 取平均分
        int sum = 0;
        for(std::deque<int>::iterator dit = d.begin(); dit != d.end(); dit++){
            sum += *dit;
        }
        int avg = sum / d.size();
        it->score = avg;
    }
}

void showScore(std::vector<Person> &v){
    std::cout << "Final Score: " << std::endl;
    for(std::vector<Person>::iterator it = v.begin(); it != v.end(); it++){
        std::cout << "name: " << (*it).name << ", socre: " << (*it).score << std::endl;
    }
}
int main(int argc, char* argv[]){

    srand( (unsigned int)time(NULL));
    // 创建 5 名选手
    std::vector<Person>v;
    createPerson(v);

    // 给选手打分
    setScore(v);

    // 显示最终得分
    showScore(v);
    return 0;
}

2--员工分组

2-1--案例分析

        公司招聘 10 个员工（ABCDEFGHIJ），员工信息包括姓名和工资，需要将员工分配到策划、美术和研发三个部门；通过 multimap 容器进行员工信息的插入，并分部分显示员工的信息；

2-2--代码实现

#include "iostream"
#include "string"
#include "vector"
#include "map"
#include "ctime"

class Person{
public:
    std::string Name;
    int Salary;
};

void CreateWorker(std::vector<Person> &v){
    std::string nameseed = "ABCDEFGHIJ";
    for(int i = 0; i < 10; i++){
        Person p;
        p.Name = "Worker";
        p.Name += nameseed[i];
        p.Salary = rand() % 10000 + 10000; // 10000 ~ 19999
        v.push_back(p);
    }
}

void SetGroup(std::vector<Person> &v, std::multimap<int, Person> &m){
    for(std::vector<Person>::iterator it = v.begin(); it != v.end(); it++){
        int ID = rand() % 3; // 0, 1, 2
        m.insert(std::make_pair(ID, *it));
    }
}

void ShowWorkerByGroup(std::multimap<int, Person> &m){
    std::cout << "Group of design: " << std::endl;
    std::multimap<int, Person>::iterator pos = m.find(0);
    int count = m.count(0);
    for(int i = 0; pos != m.end() && i < count; pos++, i++){
        std::cout << "Group: " << pos->first << " Name: " << pos->second.Name << " Salary: " << pos->second.Salary << std::endl;
    }

    std::cout << "Group of Art: " << std::endl;
    pos = m.find(1);
    count = m.count(1);
    for(int i = 0; pos != m.end() && i < count; pos++, i++){
        std::cout << "Group: " << pos->first << " Name: " << pos->second.Name << " Salary: " << pos->second.Salary << std::endl;
    }

    std::cout << "Group of research: " << std::endl;
    pos = m.find(2);
    count = m.count(2);
    for(int i = 0; pos != m.end() && i < count; pos++, i++){
        std::cout << "Group: " << pos->first << " Name: " << pos->second.Name << " Salary: " << pos->second.Salary << std::endl;
    }
}

int main(int argc, char* argv[]){
    srand((unsigned int) time(NULL));
    std::vector<Person> v;
    CreateWorker(v);

    std::multimap<int, Person> m;
    SetGroup(v, m);
    ShowWorkerByGroup(m);
    return 0;
}