202504291653 1012 The Best Rank

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks — that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90
310102     70 95 88 84
310103     82 87 94 88
310104     91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (≤2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output Specification:

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output N/A.

Sample Input:

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output:

1 C
1 M
1 E
1 A
3 A
N/A

Solve

三个科目和一个均分作为一个学生的成绩情况，题目要求查询学生每个科目以及平均分在所有学生相同类型的成绩下的排名和对应类型。

其中每个成绩类型排名时，相同成绩的按相同排名，同一个学生存在两个或多个成绩类型均位于相同名次时，按照ACME优先级，即平均分>计算机>数学>英语。

在输出前将查询和对应输出一一映射一下，做索引优化，避免每次都需要遍历整个列表。

#include<iostream>
#include<vector>
#include<algorithm>
#include<unordered_map>
#include<climits>
using namespace std;
 
const string toRankType = "ACME";
 
struct Student {
    string id;
    int score[4] = {0};
    int bestRank = INT_MAX;
    char bestRankType = ' ';
};
 
int main() {
 
    int n, m;
    cin >> n >> m;
    vector<Student> students(n);
    for(int i = 0; i < n;i ++) {
        auto &t = students[i];
        cin >> t.id >> t.score[1] >> t.score[2] >> t.score[3];
        t.score[0] = (t.score[1] + t.score[2] + t.score[3]) / 3;
    }
 
    for(int i = 0; i < 4; i++) {
        sort(students.begin(), students.end(), [i](const auto& a, const auto& b) {
            return a.score[i] > b.score[i];
        });
 
        int rank = 1;
        for(int j = 0; j < n;j ++) {
            // 处理并列排名
            if(j > 0 && students[j].score[i] != students[j-1].score[i]) 
                rank = j + 1;
            
            // 更新最佳排名
            if(rank < students[j].bestRank) {
                students[j].bestRank = rank;
                students[j].bestRankType = toRankType[i];
            }
        }
    }
 
    unordered_map<string, Student> result;
    for(auto &t : students) 
        result[t.id] = t;
 
    while(m--) {
        string s;
        cin >> s;
 
        if(result.count(s))
            cout << result[s].bestRank << " " << result[s].bestRankType << endl;
        else
            cout << "N/A" << endl;
    }
 
    return 0;
}