PAT日志 1080

顽强的小白

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 GE​​, and the interview grade GI​​. 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 GE. 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 iii-th integer is the quota of the iii-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 GE​​ and GI​​, 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

题目解析

根据学生的成绩和志愿顺序，学校的招生名额来录取学生。

• 总成绩高低排序，总成绩相同就按照GE成绩排序
• 每个学校名额有限，招完为止，但是如果最后一名有并列的可以破格录取

思路：

• 使用结构体：学生需要记录他们自己的id，因为一排序就乱了，成绩以及排名信息（方便找并列），学校不需要记录id，但是最好记录最后一名录取的学生id（方便验证是否并列）
• 排序：有两次排序，学生信息读取后需要排序。每个学校录取结束后需要把学生id从小到大排序（也考虑了用set自动排序）
• 我在这道题上栽跟头的重点来了：关于排序后排名的问题，我因为排rank的时候用的是结构体的下标，此时已经和它的id不同步了，这就影响到处理录取学生的问题，录取当前学生时，需要记录它的代号，到底是记录下标还是记录id呢？稍微一想错就会过不了测试点。因为比较是否并列用的是学生下标，因此lastID我需要记录的是下标，但是学校的录取学生名单上需要记录的是id，为了方便输出。（也有些思路稍微改改就不一样了，总之这种细节还是需要我长长心）

代码实现

const int maxm=105;
const int maxk=5;
struct student{
 int id,ge,gi,sum,rank;
 int sch[maxk];
}stu[maxn];
struct school{
 int stu[maxn];
 int stuNum;
 int quota;
 int lastId;
}sch[maxm];bool cmp(student a,student b){
 if(a.sum!=b.sum){
  return a.sum>b.sum;
 }else{
  return a.ge>b.ge;
 }
}
bool cmp1(int a,int b){
 return a<b;
}int main(){
 int n,m,k;
 scanf("%d%d%d",&n,&m,&k);
 for(int i=0;i<m;++i){
  sch[i].lastId=-1;
  scanf("%d",&sch[i].quota);
  sch[i].stuNum=0;
 }
 for(int i=0;i<n;++i){
  stu[i].id=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].sch[j]);
  }
 }
 sort(stu,stu+n,cmp);
 for(int i=0;i<n;++i ){
  if(i>0&&stu[i].ge==stu[i-1].ge&&stu[i].sum==stu[i-1].sum){
   stu[i].rank=stu[i-1].rank;
  }else {
   stu[i].rank=i;
  }
 }
 for(int i=0;i<n;++i){
  for(int j=0;j<k;++j){
   int cho=stu[i].sch[j];
   int last=sch[cho].lastId;
   int stuNum=sch[cho].stuNum;
   if(sch[cho].quota>stuNum||(last!=-1&&stu[i].rank==stu[last].rank)){
    sch[cho].stu[stuNum]=stu[i].id;
    sch[cho].stuNum++;
    sch[cho].lastId=i;
    break;
   }
  }
 }
 for(int i=0;i<m;++i){
  int stuNum=sch[i].stuNum;
  if(stuNum>0){
   sort(sch[i].stu,sch[i].stu+stuNum,cmp1);
   for(int j=0;j<stuNum;++j){
    if(j!=0)printf(" ");
    printf("%d",sch[i].stu[j]);
   }
  } 
  printf("\n");
 }
 return 0;
}