【PAT甲级】1080 Graduate Admission

本文介绍了一个研究生招生模拟算法,通过排序和模拟流程,解决了毕业生申请研究生学校的问题。算法首先按成绩排序,然后按照志愿和学校招生规则进行录取。

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It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admission procedure.

Each applicant will have to provide two grades: the national entrance exam grade G​E​​, and the interview grade G​I​​. The final grade of an applicant is (G​E​​+G​I​​)/2. The admission rules are:

  • The applicants are ranked according to their final grades, and will be admitted one by one from the top of the rank list.

  • If there is a tied final grade, the applicants will be ranked according to their national entrance exam grade G​E​​. If still tied, their ranks must be the same.

  • Each applicant may have K choices and the admission will be done according to his/her choices: if according to the rank list, it is one's turn to be admitted; and if the quota of one's most preferred shcool is not exceeded, then one will be admitted to this school, or one's other choices will be considered one by one in order. If one gets rejected by all of preferred schools, then this unfortunate applicant will be rejected.

  • If there is a tied rank, and if the corresponding applicants are applying to the same school, then that school must admit all the applicants with the same rank, even if its quota will be exceeded.

Input Specification:

Each input file contains one test case.

Each case starts with a line containing three positive integers: N (≤40,000), the total number of applicants; M (≤100), the total number of graduate schools; and K (≤5), the number of choices an applicant may have.

In the next line, separated by a space, there are M positive integers. The i-th integer is the quota of the i-th graduate school respectively.

Then N lines follow, each contains 2+K integers separated by a space. The first 2 integers are the applicant's G​E​​ and G​I​​, respectively. The next K integers represent the preferred schools. For the sake of simplicity, we assume that the schools are numbered from 0 to M−1, and the applicants are numbered from 0 to N−1.

Output Specification:

For each test case you should output the admission results for all the graduate schools. The results of each school must occupy a line, which contains the applicants' numbers that school admits. The numbers must be in increasing order and be separated by a space. There must be no extra space at the end of each line. If no applicant is admitted by a school, you must output an empty line correspondingly.

Sample Input:

11 6 3
2 1 2 2 2 3
100 100 0 1 2
60 60 2 3 5
100 90 0 3 4
90 100 1 2 0
90 90 5 1 3
80 90 1 0 2
80 80 0 1 2
80 80 0 1 2
80 70 1 3 2
70 80 1 2 3
100 100 0 2 4

Sample Output:

0 10
3
5 6 7
2 8

1 4

题目大意

这题题目有点长,主要讲的大意就是毕业季每个有n个应届生,有m个招研究生的学校,每个学生有k个志愿选择。接着给出了m个学校的招生满额人数。然后给出了n行,每行包括这个应届生的两种成绩,以及该学生的k个志愿。接着按照排序规则进行排名,然后按照排名先后进行选择志愿。最后输出每个学校的招生学生信息。

排序规则是:以grade_e+grade_i从大到小为第一优先级,如果相等则按照grade_e从大进行排序,如果都相等则排名相同。

招生规则:

  • 按照排名从前往后依次进行志愿选择
  • 如果第一志愿招生已满,则进行第二志愿、第三志愿...直到被录取,如果都没录取,那该生就很倒霉了。
  • 如果当前志愿已经招满人了,但是之前有和当前学生同排名的学生进入了目标院校,则即便已经招满人了,该生也能进入。

个人思路

还是排序算法的应用,招生过程就按照上述的招生规则进行模拟即可。

代码实现

#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <cmath>
#include <algorithm>
#include <iostream>
#define ll long long
#define eps 1e-8
#define INF 0x7FFFFFFF

using namespace std;

const int maxm = 105;
const int maxn = 40005;

// 应届生信息结构体
struct Application {
    int id, grade_e, grade_i, rank, school;
    int choice[5];
};

// 应届生排序函数 优先级:(grade_e+grade_i) / 2 > grade_e
bool cmp_app(Application a1, Application a2) {
    if (a1.grade_e+a1.grade_i != a2.grade_e+a2.grade_i)
        return a1.grade_e+a1.grade_i > a2.grade_e+a2.grade_i;
    else
        return a1.grade_e > a2.grade_e;
}

// 学校新收研究生排序,按照id从小到大
bool cmp_member(int a1, int a2) {
    return a1 < a2;
}

int quota[maxm]; // 每个学校满额数量
Application apps[maxn]; // 应届生数组
int admitted[maxm] = {0}; // 每个学校当前接受研究生的数量
vector<int> member[maxm]; // 每个学校当前接受研究生的信息

int main() {
    // 输入
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < m; i ++) {
        cin >> quota[i];
    }
    for (int i = 0; i < n; i ++) {
        apps[i].id = i;
        cin >> apps[i].grade_e >> apps[i].grade_i;
        for (int j = 0; j < k; j ++) {
            cin >> apps[i].choice[j];
        }
    }
    // 排序
    sort(apps, apps+n, cmp_app);
    // 排名
    apps[0].rank = 1;
    for (int i = 1; i < n; i ++) {
        if (apps[i].grade_e == apps[i-1].grade_e && apps[i].grade_i == apps[i-1].grade_i)
            apps[i].rank = apps[i-1].rank;
        else
            apps[i].rank = i+1;
    }
    // 按照排名后顺序进行目标院校选择
    for (int i = 0; i < n; i ++) {
        Application a_tmp = apps[i];
        for (int j = 0; j < k; j ++) {
            int choosed = a_tmp.choice[j];
            // 如果目标院校没收满
            if (admitted[choosed] < quota[choosed]) {
                admitted[choosed] ++;
                member[choosed].push_back(a_tmp.id);
                apps[i].school = choosed;
                break;
            }
            // 目标院校已经收满
            else {
                bool flag = false; // 判断是否有同名次的人在这个学校
                int pos = i-1;
                while (a_tmp.rank == apps[pos].rank) {
                    if (apps[pos].school == choosed) {
                        flag = true;
                        break;
                    }
                    pos --;
                }
                // 如果有同名次的人在这个学校该生也能进
                if (flag) {
                    admitted[choosed] ++;
                    member[choosed].push_back(a_tmp.id);
                    apps[i].school = choosed;
                    break;
                }
            }
        }
    }
    // 输出
    for (int i = 0; i < m; i ++) {
        if (admitted[i] != 0) {
            sort(member[i].begin(), member[i].end(), cmp_member);
            for (int j = 0; j < member[i].size(); j ++) {
                cout << member[i][j];
                if (j != member[i].size()-1) cout << " ";
            }
        }
        cout << endl;
    }
    return 0;
}

总结

学习不息,继续加油

 

逻辑斯蒂回归(Logistic Regression)是一种常用的分类算法,常用于二分类问题中,如预测研究生能否被录取。Kaggle的Graduate Admission数据集包含了申请人的各项信息,例如GRE分数、TOEFL分数、大学GPA、科研经验、推荐信等,目标变量通常是“是否被录取”(是否被研究生院接受)。 首先,我们来理解数据集属性的意义: 1. GRE Score: 研究生入学考试成绩 2. TOEFL Score: 英语水平测试得分 3. University Rating: 学校排名 4. SOP: Statement of Purpose(个人陈述)的质量 5. LOR: Letter of Recommendation(推荐信)的质量 6. CGPA: 学术平均绩点 7. Research: 科研经历(0或1) 8. Chance of Admit: 录取概率(这个不是原始数据,而是我们最终需要预测的目标) 数据预处理步骤主要包括: 1. **加载数据**:使用pandas库读取csv文件并查看基本信息。 2. **缺失值处理**:检查是否存在缺失值,并选择填充、删除或估算策略。 3. **编码分类变量**:将类别型特征转换成数值型,如使用one-hot encoding或者LabelEncoder。 4. **标准化或归一化**:对于数值型特征,通常会做数据缩放,如Z-score标准化或min-max归一化。 5. **划分训练集和测试集**:通常采用80%的数据作为训练集,剩余的20%作为测试集。 6. **特征工程**:如果有必要,可以创建新的特征或调整现有特征。 逻辑斯蒂回归的预测原理是基于sigmoid函数,该函数将线性组合后的输入映射到0到1之间,表示事件发生的可能性。模型学习如何调整权重系数,使得给定输入条件下,正类(如录取)的概率最大化。 实现过程(Python示例,假设使用sklearn库): ```python import pandas as pd from sklearn.model_selection import train_test_split from sklearn.preprocessing import StandardScaler from sklearn.linear_model import LogisticRegression from sklearn.metrics import accuracy_score # 1. 加载数据 data = pd.read_csv('Admission_Predict.csv') # 2. 数据预处理 # ... 缺失值处理、编码、标准化等操作 # 3. 划分特征和目标 X = data.drop('Chance of Admit', axis=1) y = data['Chance of Admit'] # 4. 划分训练集和测试集 X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42) # 5. 特征缩放 scaler = StandardScaler() X_train_scaled = scaler.fit_transform(X_train) X_test_scaled = scaler.transform(X_test) # 6. 创建模型并拟合 model = LogisticRegression() model.fit(X_train_scaled, y_train) # 7. 预测 y_pred = model.predict(X_test_scaled) # 8. 评估模型性能 accuracy = accuracy_score(y_test, y_pred) print(f"Accuracy: {accuracy}") ```
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