pat 1080. Graduate Admission (30) 浙大复试上机第四题

1080. Graduate Admission (30)

时间限制

200 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

提交代码

思路：

1:学生按照指定的方式进行排序，排序优先级为最终分数高->ge高->，两个都一样，则相等。

2:然后根据排好序的学生次序，对每个学生指定其排名，注意这个排名是相同等级的排名一样，比如   (90,90),(90,90),(89,91),排名为1,1,2;

3:为学生指定排名，主要是因为对于学校，如果有一个学生恰好进入了这个学校，那么其他和这个学生排名相同的学生，只要选了这个学校，都有资格进入，

不收学校招生人数的影响。

4:然后就是从高到低，一次查看学生的志愿。每次学生申请学校成功，都要把其排名告诉给学校。这样，学校在招生满的时候，如果收到一个排名和其最低

招生排名相同的学生，就可以录取这个学生了。

5:接着就是打印每个学校录取的学生的id，排序从低到高输出就行。不要输出多余的信息。

//15:17 reading-> 15:23 thinking ->15:28 coding->16:20 finished
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

struct Students
{
  int id;
  double g1;
  double g2;
  double fg;
  int rank;
  int apps[7];
};
struct School
{
  int quota;
  vector<int> stus;
  int lowerRank;
};
Students stus[40003];
School schs[103];
int cmp(const void *a,const void *b)
{
  Students x = *((Students *)a);
  Students y = *((Students *)b);
  if(x.fg!=y.fg)
  {
    return (x.fg>y.fg)?-1:1;
  }else if(x.g1!=y.g1)
  {
    return (x.g1>y.g1)?-1:1;
  }else
  {
    return 0;
  }
}

int main()
{
  int n,m,k;
  int i,j,qu,s;
  scanf("%d%d%d",&n,&m,&k);
  for(i=0;i<m;i++)
  {
    scanf("%d",&schs[i].quota);
    schs[i].lowerRank = -1;
  }
  for(i=0;i<n;i++)
  {
    scanf("%lf%lf",&stus[i].g1,&stus[i].g2);
    stus[i].fg = (stus[i].g1+stus[i].g2)/2;
    stus[i].id = i;
    for(j=0;j<k;j++)
    {
      scanf("%d",&s);
      stus[i].apps[j] = s;
    }
  }
  qsort(&stus[0],n,sizeof(Students),cmp);
  int r=1,same = 0;
  i = 0;
  while(true)
  {
    same = 0;
    if(i>=n)
    {
      break;
    }
    stus[i].rank = r;
    while(i<n-1&&stus[i].fg==stus[i+1].fg && 
    stus[i].g1==stus[i+1].g1)
    {
      stus[i+1].rank = r;
      same++;
      i++;
    }
    r += same;
    r++;
    i++;
  }
  for(i=0;i<n;i++)
  {
    for(j=0;j<k;j++)
    {
      s = stus[i].apps[j];
      if(schs[s].stus.size()<schs[s].quota)
      {
        schs[s].stus.push_back(stus[i].id);
        schs[s].lowerRank = stus[i].rank;
        break;
      }else
      {
        if(schs[s].lowerRank==stus[i].rank)
        {
          schs[s].stus.push_back(stus[i].id);
          break;
        }
      }
    }
  }
  for(i=0;i<m;i++)
  {
    bool flag = false;
    if(schs[i].stus.size()>0)
    {
      //printf("%d ",i);
      sort(schs[i].stus.begin(),schs[i].stus.end());
      for(j=0;j<schs[i].stus.size();j++)
      {
        printf("%d",schs[i].stus[j]);
        if(j!=schs[i].stus.size()-1)
        {
          printf(" ");
        }
      }
    }else
    {
      printf("\n");
      flag = true;
    }
    if(i!=m-1&&!flag)
    {
      printf("\n");
    }
  }
  return 0;
}