1080 Graduate Admission (30 分)-CFANZ编程社区

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 (GE+GI)/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 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 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:

难点：生词太多看不懂，好多单词百度的。。不知道考场上能不能做出来。。。。
因为30分，就一直很谨慎。。。结果运气好，一次过。。但是花费的时间太多。
30分答题就是烦。。不会特别为难你。（数据很弱）

N 总申请数(志愿数)M 大学数K 志愿书有几个选择
第二行M个数分别表示M个学校的指标要前多少名的并列的必须收没有人报的可以缺招有报的不管多差按排名录取
G1 G2 1 2 3 //两个成绩三个prefer的学校
ve.back()获取vector最后一个元素别忘记括号

#include<bits/stdc++.h>
using namespace std;

struct student{
  int G1,G2,sum;
  int school[5];
  int order,rank;//申请序号和排名 
}stu[40010];
int N,M,K;
int quota[110];

bool compare(student a,student b){
  if(a.sum!=b.sum) return a.sum>b.sum;
  if(a.G1!=b.G1) return a.G1>b.G1;
  return a.order<b.order;//自己加的 简化输出逻辑   
}
bool compare1(student a,student b){
  return a.order<b.order;
}

int main(){
  //freopen("in.txt","r",stdin);
  cin>>N>>M>>K;
  for(int i=0;i<M;i++) cin>>quota[i];//录取 志愿填我们学校的多少位   除非没人填  否则再差也要 
  for(int i=0;i<N;i++){
    cin>>stu[i].G1>>stu[i].G2;
    stu[i].sum=stu[i].G1+stu[i].G2;//计算总分
    stu[i].order=i;//保存原始序号 
    for(int j=0;j<K;j++){
      cin>>stu[i].school[j];
    }
  }
  
  sort(stu,stu+N,compare);
  for(int i=0;i<N;i++){
    stu[i].rank=i+1;
    if(i>0&&stu[i].sum==stu[i-1].sum&&stu[i].G1==stu[i-1].G1) stu[i].rank=stu[i-1].rank;//并列名次 
  } 
  
  
  //计算录取情况  遍历高分学生比较好  因为志愿在前的会被抢走
  vector<student> admin[110];//admin[i].push_back(n)  第i号学校录取了n号学生
  for(int i=0;i<N;i++){
    for(int j=0;j<K;j++){
      int n=stu[i].school[j];
      if(quota[n]>0){
        admin[n].push_back(stu[i]);
        quota[n]--; 
        break;//一旦被录取了 就立刻停止找学校 
      }else if(admin[n].back().rank==stu[i].rank){//ve.back()获取vector最后一个元素 
        //排名相同 破格录取
        admin[n].push_back(stu[i]); 
        break;//一旦被录取了 就立刻停止找学校 
      }
    }
  } 
  
  for(int i=0;i<M;i++){
    sort(admin[i].begin(),admin[i].end(),compare1);
    for(int j=0;j<admin[i].size();j++){
      if(j!=0) cout<<" ";
      cout<<admin[i][j].order;
    }
    cout<<endl;
  }
  return 0;
}