PAT(A) 1080. Graduate Admission (30)

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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=40010;

struct Student{
    int GE, GI, sum;    //初试成绩，面试成绩，成绩总合
    int r, stuID;       //总排名，考号（stu[]按成绩排序前的初始编号）设：排序后的称为编号
    int choice[6];      //k个志愿学校的编号
}stu[maxn];

struct School{
    int quota;          //该校招生人数总额
    int stuNum;         //当前实际招生人数
    int id[maxn];       //招收的考生编号
    int lastAdmint;     //记录最后一个招收的考生编号（即实际招收的考生数）
}sch[110];

bool cmpStu(Student a, Student b){
    if(a.sum != b.sum)  return a.sum>b.sum; //按总分从高到低排序
    else    return a.GE>b.GE;               //总分相同，按GE从高到低排序
}
bool cmpID(int a, int b){
    return stu[a].stuID < stu[b].stuID;     //按考生初始ID从小到大排序
}

int main()
{
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);       //考生人数，学校数，每人可申请的学校数
    //初始化每个学校
    for(int i=0; i<m; i++){
        scanf("%d", &sch[i].quota);    //输入招生人数总额度
        sch[i].stuNum=0;               //当前实际招生人数初始化为0
        sch[i].lastAdmint=-1;  //最后一个招收的学生编号初始化为-1，表示不存在（未招收）
    }
    //初始化每个考生
    for(int i=0; i<n; i++){
        stu[i].stuID=i;                             //考生初始ID为i (0->n-1)
        scanf("%d%d", &stu[i].GE, &stu[i].GI);      //初试成绩，面试成绩
        stu[i].sum = stu[i].GE+stu[i].GI;           //总成绩
        for(int j=0; j<k; j++)
            scanf("%d", &stu[i].choice[j]);         //该生申请的k个学校的编号
    }
    //给n个考生按成绩排序
    sort(stu, stu+n, cmpStu);

    //计算每个考生的排名
    for(int i=0; i<n; i++){ //若该考生与前面一个考生总分和初试成绩都相同，则排名也相同
        if(i>0 && stu[i].sum==stu[i-1].sum && stu[i].GE==stu[i-1].GE)
            stu[i].r=stu[i-1].r;
        else
            stu[i].r=i;
    }
    //枚举每个考生i，判断其会被哪所学校录取 （此处及以下的i为考生编号）
    for(int i=0; i<n; i++){
        for(int j=0; j<k; j++){  　　　　　　　　//枚举考生i的k个志愿学校
            int cho = stu[i].choice[j]; 　　　　//暂存考生i的第j个志愿学校的编号
            int num = sch[cho].stuNum; //志愿学校当前的实际招生人数（下一个录取的考生编号从num开始
            int last = sch[cho].lastAdmint; 　 //志愿学校最后一个录取的考生编号
            //如果人数未满 or 该校最后一个录取的考生stu[last]存在
            //即该校有招生:最后一个录取的考生编号!=-1
            //且与当前待判定的考生stu[i]排名相同
            if( num<sch[cho].quota || (last!=-1 && stu[i].r==stu[last].r) ){
                //录取该考生
                sch[cho].id[num]=i;    //将第一次排序后各考生的编号记录在 sch[cho].id[num]中
                sch[cho].lastAdmint=i; //该学校最后一个录取的考生变为i
                sch[cho].stuNum++;     //当前招生人数+1
                break;
            }
        }
    }
    //对m所学校的招生情况进行输出
    for(int i=0; i<m; i++){
        if(sch[i].stuNum>0){    //如果有招到学生
            //在sch[i].id[stuNum]即该校招入的学生范围内，按这些学生的初始ID从小到大排序
            sort(sch[i].id, sch[i].id+sch[i].stuNum, cmpID);
            //输出该校招入学生的初始ID
            for(int j=0; j<sch[i].stuNum; j++){
                //sch[i].id[j]为考生编号（排序后）
                //stu[sch[i].id[j]].stuID为考号（排序前的初始编号）
                printf("%d", stu[sch[i].id[j]].stuID);
                if(j<sch[i].stuNum-1)   //格式输出
                    printf(" ");
            }
        }
        printf("\n");
    }
    return 0;
}