PAT甲级1012 The Best Rank

重复代码很多的一道模拟题，感觉写得不够好。

坑点就是，可能会忘记分数相等时，排名是并列的，取高。至于课程的优先级问题，把高优先级的课程放在低优先级之后排序即可，如果排名相等，高优先级会覆盖掉低优先级。

然后，最后针对 m 个 id，输出最高排名。如果用线性扫描，复杂度 O(nm)，也不会超时。可以考虑对 id 排序，然后二分，我就懒得写了。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3+3;
const double EPS = 1e-6;
char course[4] = {'A', 'C', 'M', 'E'};
struct Student {
    string id;
    double mark[4];
    int bestRank;
    char bestCourse;

    Student() {
        bestRank = maxn;
    }

    void getInfo() {
        cin >> id >> mark[1] >> mark[2] >> mark[3];
    }

    void getAve() {
        mark[0] = (mark[1] + mark[2] + mark[3]) / 3.0;
    }
} students[maxn];
int n, m;

void read() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        students[i].getInfo();
        students[i].getAve();
    }
}

bool cmpA(const Student& s1, const Student& s2) {
    return s1.mark[0] - s2.mark[0] > EPS;
}

bool cmpC(const Student& s1, const Student& s2) {
    return s1.mark[1] - s2.mark[1] > EPS;
}

bool cmpM(const Student& s1, const Student& s2) {
    return s1.mark[2] - s2.mark[2] > EPS;
}

bool cmpE(const Student& s1, const Student& s2) {
    return s1.mark[3] - s2.mark[3] > EPS;
}

void getBestRank(int c) {
    int high = maxn;
    for (int i = 1; i <= n; ++i) {
        // 分数有可能相同，排名并列；考虑到精度，直接用==就行（逃
        if (students[i-1].mark[c] == students[i].mark[c]) {
            high = min(high, i-1);
        } else {
            high = i;
        }

        if (students[i].bestRank >= high) {
            students[i].bestRank = high;
            students[i].bestCourse = course[c];
        }
    }
}

void solve() {
    sort(students+1, students+1+n, cmpE);
    getBestRank(3);
    sort(students+1, students+1+n, cmpM);
    getBestRank(2);
    sort(students+1, students+1+n, cmpC);
    getBestRank(1);
    sort(students+1, students+1+n, cmpA);
    getBestRank(0);

    string id;
    while (m--) {
        cin >> id;
        bool isExist = false;
        for (int i = 1; i <= n; ++i) {
            if (students[i].id == id) {
                cout << students[i].bestRank << " " << students[i].bestCourse << endl;
                isExist = true;
                break;
            }
        }
        if (!isExist) {
            cout << "N/A" << endl;
        }
    }
}

int main() {
    read();
    solve();
    return 0;
}