题解 | 考试招生

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 1005;
const int M = 105;

struct Stu {
    int id;
    int scores[4];
    int total;
    bool admitted;
};

Stu stus[N];
int n, m, k, c;

bool compare(int cand_a, int cand_b) {
    const Stu& s1 = stus[cand_a];
    const Stu& s2 = stus[cand_b];

    // 总分不相同按总分排序
    if(s1.total != s2.total) return s1.total > s2.total;

    // 第c门不为0，按第c门排序
    if(c > 0 && s1.scores[c - 1] != s2.scores[c - 1]) return s1.scores[c - 1] > s2.scores[c - 1];

    // 都相等，按id大小排序
    return s1.id < s2.id;
}

int main() {
    cin >> n >> m;
    // 读入所有学生分数，计算总分
    for (int i = 0; i < n; i++) {
        stus[i].id = i + 1;
        stus[i].total = 0;
        stus[i].admitted = false;
        for (int j = 0; j < 4; j++) {
            cin >> stus[i].scores[j];
            stus[i].total += stus[i].scores[j];
        }
    }

    // 处理各系的要求
    for (int d = 0; d < m; d++) {
        cin >> k >> c;
        int cand[N], count = 0;

        // 收集没被录取的的学生
        for (int i = 0; i < n; i++)
            if (!stus[i].admitted) cand[count++] = i;
        sort(cand, cand + count, compare);

        // 确定实际录取人数
        int actual_k = (k <= count) ? k : count;

        if (actual_k > 0) {
            int last_idx = cand[actual_k - 1];
            int last_total = stus[last_idx].total;
            int last_score = (c > 0) ? stus[last_idx].scores[c - 1] : 0;

            // 检查同分情况
            for (int i = actual_k; i < count; i++) {
                int curr_idx = cand[i];

                // 检查总分是否相同
                if (stus[curr_idx].total != last_total) break;

                // 如果c>0，检查第c门成绩是否相同
                if (c > 0 && stus[curr_idx].scores[c - 1] != last_score) break;
                actual_k++;
            }
        }

        // 输出结果
        cout << actual_k << " ";
        for(int i = 0; i < actual_k; i++) {
            int idx = cand[i];
            stus[idx].admitted = true;
            cout << stus[idx].id << " ";
        }
        cout << endl;
    }
}