1080. Graduate Admission (30)

题目链接：http://www.patest.cn/contests/pat-a-practise/1080

题目：

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

分析：

有两个分数：Ge和Gi，按照它们的平均数(Ge+Gi)/2来评分，如果平均分相同的按照Ge来排列，如果都相同，则他们的排名相同

如果有分数相同的学生，则学校及时超过配额也要全部接收

对学生进行排序，按照分数高低进行选择

最后学校输出的结果还得按照学生id的顺序排好。

AC代码：

#include
#include
#include
#include
using namespace std;
struct Student{
 int rank;
 int id;
 int Ge;
 int Gi;
 float average;
 vectorapp_School;
 Student(){
  rank = -1;
  id = -1;
  Ge = 0;
  Gi = 0;
  average = 0;
  app_School.clear();
 }
};
struct School{
 int id;
    int quota;
 vectorreceive_Stu;
 School(){
  id = -1;
  quota = 0;
  receive_Stu.clear();
 }
};
bool cmp(Student A, Student B){
 if (A.average != B.average)return A.average > B.average;
 else if (A.Ge != B.Ge)return A.Ge > B.Ge;
 else return A.id < B.id;
}
bool cmp_by_id(int A,int B){
 return A < B;
}
int main(){
 freopen("F://Temp/input.txt", "r", stdin);
 int N, M, K;
 cin >> N >> M >> K;
 School *school = new School[M];
 Student *student = new Student[N];
 for (int i = 0; i < M; i++){
  school[i].id = i;
  cin >> school[i].quota;
 }
 for (int i = 0; i < N; i++){
  student[i].id = i;
  cin >> student[i].Ge >> student[i].Gi;
  student[i].average = (student[i].Ge*1.0 + student[i].Gi) / 2;
  for (int j = 0; j < K; j++){
   int app_school_input;
   cin >> app_school_input;
   student[i].app_School.push_back(app_school_input);
  }
 }
 sort(student, student + N, cmp);
 for (int i = 0; i < N; i++){
  if (i == 0){ student[i].rank = 0; }
  else{
   if (student[i].average != student[i - 1].average || student[i].Ge != student[i - 1].Ge){
    student[i].rank = i;
   }
   else{
    student[i].rank = student[i - 1].rank;
   }
  }
 }
 int *id2rank = new int[N]; //易错点，因为前后的编号变化了
 for (int i = 0; i < N; i++){
  id2rank[student[i].id] = student[i].rank;
 }
 for (int i = 0; i < N; i++){
  for (int j = 0; j < K; j++){
   int cur_school = student[i].app_School[j];
   if (school[cur_school].receive_Stu.size() < school[cur_school].quota){
    //如果学校的配额还没满，则接收
    school[cur_school].receive_Stu.push_back(student[i].id);
    break;
   }
   else if(school[cur_school].quota > 0){ //这里也要注意，怕万一学校的quota为0
    int size = school[cur_school].receive_Stu.size();
    int last_student = school[cur_school].receive_Stu[size - 1];
    if (student[i].rank == id2rank[last_student]){
     //如果学校的配额满了，但是同学的平均和Ge分数都和学校最后一位同学相同，也接收
     school[cur_school].receive_Stu.push_back(student[i].id);
     break;
    }
    else{
     continue;
    }
   }
  }
 }
 for (int i = 0; i < M; i++){
  int size = school[i].receive_Stu.size();
  sort(school[i].receive_Stu.begin(), school[i].receive_Stu.end(), cmp_by_id);
  for (int j = 0; j < size; j++){
   if (j == 0)cout << school[i].receive_Stu[j];
   else{
    cout << " " << school[i].receive_Stu[j];
   }
  }
  cout << endl;
 }
 return 0;
}

截图：

——Apie陈小旭