1080. Graduate Admission (30)

Posted 2020-10-04

1080. Graduate Admission (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

It is said that in 2013, there were 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

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int n, m, k;
vector<int> school_quota;
vector<vector<int>> school_admit;
int adm_rank[101];
struct STU
{
    int index;
    int GE, Gi;
    int rank;
    int appli[5];
    int adm;
    STU():adm(-1){}
}stu[40001];
bool cmp(STU a, STU b)
{
    int fa = (a.GE + a.Gi) / 2;
    int fb = (b.GE + b.Gi) / 2;
    if (fa != fb)
    {
        return fa > fb;
    }
    else return a.GE > b.GE;
}
int main()
{
    cin >> n >> m >> k;
    int temp;
    for (int i = 0; i < m; i++)
    {
        cin >> temp;
        school_quota.push_back(temp);
    }
    for (int i = 0; i < n; i++)
    {
        cin >> stu[i].GE >> stu[i].Gi;
        for (int j = 0; j < k; j++)
            cin >> stu[i].appli[j];
        stu[i].index = i;
    }
    sort(stu, stu + n, cmp);
    stu[0].rank = 0;
    for (int i = 1; i < n; i++)
    {
        if (stu[i].GE  == stu[i - 1].GE &&stu[i].Gi == stu[i - 1].Gi)
            stu[i].rank = stu[i - 1].rank;
        else stu[i].rank = i;
    }
    school_admit.resize(m);
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < k; j++)
        {
            if (school_quota[stu[i].appli[j]] > 0)
            {
                if (stu[i].adm != -1) break;
                school_admit[stu[i].appli[j]].push_back(stu[i].index);
                stu[i].adm = stu[i].appli[j];
                school_quota[stu[i].appli[j]]--;
                if (adm_rank[stu[i].adm] < stu[i].rank)
                    adm_rank[stu[i].adm] = stu[i].rank;
                break;
            }
            //int rank;
            //for (int a = 0; a < school_admit[stu[i].appli[j]].size(); a++)
            //{
            //    rank = stu[school_admit[stu[i].appli[j]][a]].rank;
            //}
            if (stu[i].rank <= adm_rank[stu[i].appli[j]])//曾出错，以为和学生排名的上一个比较。其实是录取的上一个，或排名最后的一个。
            {
                if (stu[i].adm != -1) break;
                school_admit[stu[i].appli[j]].push_back(stu[i].index);
                stu[i].adm = stu[i].appli[j];
                break;
            }
        }
    }
    for (int i = 0; i < m; i++)
    {
        if (school_admit[i].size())
        {
            sort(&school_admit[i][0], &school_admit[i][0] + school_admit[i].size());
            cout << school_admit[i][0];
            for (int j = 1; j < school_admit[i].size(); j++)
            {
                cout << " " << school_admit[i][j];
            }
            cout << endl;
        }
        else cout << endl;
    }
}

1 排序会破坏原始顺序，在需要id的时候加入id属性

2 适当的数据结构会大大减少代码复杂性，和出错可能。记录之前的信息。

3 在某些时候对某个关键值不要想当然，或许需要记录下来

4 低级错误，在代码中错误使用常量，而不是变量。