pat甲 A1080

1080 Graduate Admission (30 分)

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:

题目要求

本题给n个志愿申请，m个学校，每个志愿可申请k个学校，每个学校的录取名额有限给出n个申请人的初试成绩和复试成绩以及申请的k个学校，按照总成绩排名，总成绩相同的按照初试成绩排名，两者都相同的视为同一个排名，依次录取，输出每个学校的录取情况。

解法

①：将所有申请视为结构体数组，按照题目要求构造排序函数进行排名。

struct applicant
{
    int id, Ge, Gi, sumG; //下标，初试成绩，复试成绩,总成绩
    int prefer[5];        //学校志愿
    int schoolId;         //被哪个学校录取了
} Applicant[maxn];

②：对于每个学校的录取结果用vector

AC代码

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
const int maxn = 40010;
const int maxm = 110;
struct applicant
{
    int id, Ge, Gi, sumG; //下标，初试成绩，复试成绩,总成绩
    int prefer[5];        //学校志愿
    int schoolId;         //被哪个学校录取了
} Applicant[maxn];

int n, m, k;
vector<int> quota;     //各个学校的录取名额
vector<int> ans[maxm]; //每个学校真实的录取情况

bool cmp(applicant a, applicant b)
{
    if (a.sumG != b.sumG)
        return a.sumG > b.sumG;
    else
        return a.Ge > b.Ge;
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    quota.resize(m);
    for (int i = 0; i < m; i++)
    {
        scanf("%d", &quota[i]);
    }
    for (int i = 0; i < n; i++)
    {
        Applicant[i].id = i;
        scanf("%d%d", &Applicant[i].Ge, &Applicant[i].Gi);
        Applicant[i].sumG = Applicant[i].Ge + Applicant[i].Gi;
        for (int j = 0; j < k; j++)
        {
            scanf("%d", &Applicant[i].prefer[j]);
        }
    }
    sort(Applicant, Applicant + n, cmp);
    //开始录取
    for (int i = 0; i < k; i++) //第一名
    {
        int t = Applicant[0].prefer[i]; //优先录取学校
        if (quota[t] > 0)               //还有名额
        {
            ans[t].push_back(Applicant[0].id);
            Applicant[0].schoolId = t;
            quota[t]--;
            break;
        }
    }
    for (int i = 1; i < n; i++)
    {
        for (int j = 0; j < k; j++)
        {
            int t = Applicant[i].prefer[j]; //优先录取学校
            if (quota[t] > 0)
            {
                ans[t].push_back(Applicant[i].id);
                Applicant[i].schoolId = t;
                quota[t]--;
                break;
            }
            //如果两个排名相同的人录取一个学校，当学校名额为0时，照样录取
            else if (quota[t] == 0 && Applicant[i].sumG == Applicant[i - 1].sumG && Applicant[i].Ge == Applicant[i - 1].Ge && t == Applicant[i - 1].schoolId)
            {
                ans[t].push_back(Applicant[i].id);
                Applicant[i].schoolId = t;
                break;
            }
        }
    }
    for (int i = 0; i < m; i++)
    {
        sort(ans[i].begin(), ans[i].end());
        for (int j = 0; j < ans[i].size(); j++)
        {
            printf("%d", ans[i][j]);
            if (j != ans[i].size() - 1)
                printf(" ");
        }
        printf("\n");
    }
    return 0;
}

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