1080. Graduate Admission (30)

题目如下：

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

题目要求对研究生入学考试的成绩进行按照一定的规则进行排名，然后根据每个人的志愿，按照排名先后根据学校的指标进行录取。

一个关键点在于排名一致的人如果申请同一个学校，并且此时指标还有剩余，则不论是否超标都要全部录取。

最关键的是排名算法的实现，先按照总分比较，总分一致再按照笔试成绩比较，笔试成绩相同的拥有相同的排名。

为了处理超标录取的情况，每个学校都记录一个last_rank代表上次录取的人的排名，如果新申请到来时已招满但是排名等于last_rank，说明属于排名一致的人申请同一学校，也应当被录取。

代码如下：

#include 
#include 
#include 
#include 

using namespace std;

struct School{
    int quota;
    int last_rank;
    vector admitList;

    School(){
        last_rank = -1;
    }

};

struct Student{
    int num;
    int rank;
    int Ge;
    int Gi;
    int Gf;
    vector applyList;

    bool operator < (const Student &other) const{
        if(Gf > other.Gf){
            return true;
        }else if(Gf == other.Gf){
            if(Ge > other.Ge){
                return true;
            }else{
                return false;
            }
        }else{
            return false;
        }
    }

};

int main()
{
    int N,M,K;
    cin >> N >> M >> K;
    vector schs(N);
    vector stus(N);
    for(int i = 0; i < M; i++){
        scanf("%d",&schs[i].quota);
    }
    int Ge,Gi,apply;
    for(int i = 0; i < N; i++){
        scanf("%d%d",&Ge,&Gi);
        Student &stu = stus[i];
        stu.Ge = Ge;
        stu.Gi = Gi;
        stu.Gf = (Ge + Gi) / 2;
        stu.num = i;
        for(int i = 0; i < K; i++){
            scanf("%d",&apply);
            stu.applyList.push_back(apply);
        }
    }
    sort(stus.begin(),stus.end());

    int rank = 0;
    int rk_Gf = -1;
    int rk_Ge = -1;
    int rk_cnt = 1;
    int inner_cnt = 1;
    for(int i = 0; i < N; i++){
        Student &stu = stus[i];
        vector applys = stu.applyList;

        if(stu.Gf != rk_Gf){
            rk_Gf = stu.Gf;
            rk_Ge = stu.Ge;
            rank += rk_cnt;
            rk_cnt = 0;
        }else if(stu.Ge != rk_Ge){
            rk_Ge = stu.Ge;
            rank += rk_cnt;
            rk_cnt = 0;
        }
        rk_cnt++;
        stu.rank = rank;
    }
    for(int i = 0; i < N; i++){
        vector applys = stus[i].applyList;
        rank = stus[i].rank;
        for(int j = 0; j < applys.size(); j++){
            School &sch = schs[applys[j]];
            if(sch.quota > 0){
                sch.admitList.push_back(stus[i].num);
                sch.last_rank = rank;
                sch.quota--;
                break;
            }else if(sch.last_rank == rank){
                sch.admitList.push_back(stus[i].num);
                break;
            }
        }
    }
    for(int i = 0; i < M; i++){
        School &sch = schs[i];
        if(sch.admitList.size() > 0){
            sort(sch.admitList.begin(),sch.admitList.end());
            printf("%d",sch.admitList[0]);
            for(int j = 1; j < sch.admitList.size(); j++){
                printf(" %d",sch.admitList[j]);
            }
        }
        cout << endl;
    }

    return 0;
}

转载于:https://www.cnblogs.com/aiwz/p/6154037.html