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（定额） 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

参考代码：

#include
#include
using namespace std;
struct Student 
{
  int GE, GI, sum;
  int choice[6];    //K个志愿学校编号
  int r, stuID;     //排名，考生编号
}stu[40010];

struct School
{
  int quota;        //招生人数总额
  int stuNum;       //实际招收人数
  int id[40010];  //招收的考生编号
  int lastAdmit;    //记录最后一个招生考生的编号
}sch[110];

bool cmpStu(Student a, Student b){
  if(a.sum != b.sum) return a.sum > b.sum;
  return a.GE > b.GE;
} 

bool cmpID(int a, int b){
  return stu[a].stuID < stu[b].stuID;
}

int main(int argc, char const *argv[])
{
  int N,M,K;
  scanf("%d%d%d", &N, &M, &K); //考生人数，学校数，每个人可申请的学校数

  for (int i = 0; i < M; ++i)
  {
    scanf("%d", &sch[i].quota); //计划招生人数
    sch[i].stuNum = 0;          //实际招生人数初始化为0  
    sch[i].lastAdmit = -1;      //最后一个招收的学生编号为-1，表示不存在
  }

  for (int i = 0; i < N; ++i)  //初始化每个考生
  {
    stu[i].stuID = i;
    scanf("%d%d", &stu[i].GE, &stu[i].GI);
    stu[i].sum = stu[i].GE + stu[i].GI;
    for (int j = 0; j < K; ++j)
     {
       scanf("%d", &stu[i].choice[j]);
     } 
  }
  sort(stu, stu + N ,cmpStu);   //给n个考生按成绩排序  

  for (int i = 0; i < N; ++i)//计算n个考生的排名
  {
    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;
    }
  }

for (int i = 0; i < N; ++i)   //判断每个考生，被哪所学校录取了
{
  for (int j = 0; j < K; ++j)   //枚举每个考生的志愿
  {
    int choice = stu[i].choice[j];      //考生i的第j个学校编号
    int num = sch[choice].stuNum;    //选择学校的当前招生人数
    int last = sch[choice].lastAdmit;//选择学校最后一个录取考生编号
    if(num < sch[choice].quota || (last != - 1 && stu[i].r == stu[last].r)){
      sch[choice].id[num] = i;
      sch[choice].lastAdmit = i;
      sch[choice].stuNum++;
      break;
    }
  }
}

for (int i = 0; i < M; ++i) //对于m所学校
{
  if(sch[i].stuNum > 0){    //如果有招到学生
    sort(sch[i].id, sch[i].id + sch[i].stuNum, cmpID);
    for (int j = 0; j < sch[i].stuNum; ++j)
    {
      printf("%d", stu[sch[i].id[j]].stuID);
      if(j < sch[i].stuNum - 1)
        printf(" ");
    }
  }
  printf("\n");
}
  return 0;
}