PAT (Advanced Level) 1080. Graduate Admission (30) 模拟高考录取，结构体排序

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 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:

0 10
3
5 6 7
2 8

1 4

根据输入记录每个学生的id、各科分数、总分、所有志愿。根据分数进行排序后，修正每个学生的排名。

记录每个学校的录取余量和前一个录取学生的排名（初始化为0）。

按照学生的排名对所有学生进行遍历。遍历每一个学生的所有志愿，若该学生的排名与所填报学校上一个录取的学生一样，不管学校余量，直接录取，学校余量减1（可能减到负数）。若该学生的排名与学校上一个录取的学生不一样，则只在学校余量大于0的时候录取，同时学校余量减1。

/*2015.7.30cyq*/
#include <iostream>
#include <vector>
#include <fstream>
#include <algorithm>
using namespace std;

//ifstream fin("case1.txt");
//#define cin fin

struct stu{
int id;
int GE;
int GI;
int sum;
int rank;
vector<int> choice;
};

bool cmp(const stu &a,const stu &b){
if(a.sum>b.sum)
return true;
else if(a.sum==b.sum){
if(a.GE>b.GE)
return true;
}
return false;
}

int main(){
int N,M,K;
cin>>N>>M>>K;
vector<int> school(M);//招生余量
for(int i=0;i<M;i++)
cin>>school[i];

vector<stu> stus(N);
int x;
for(int i=0;i<N;i++){
stus[i].id=i;
cin>>stus[i].GE;
cin>>stus[i].GI;
stus[i].sum=stus[i].GE+stus[i].GI;
for(int j=0;j<K;j++){
cin>>x;
stus[i].choice.push_back(x);
}
}

sort(stus.begin(),stus.end(),cmp);
int rk=1;
stus[0].rank=1;
for(int i=1;i<N;i++){//与前一个学生分数不完全一样就更新排名
if(stus[i].sum!=stus[i-1].sum)
rk=i+1;
else if(stus[i].GE!=stus[i-1].GE)
rk=i+1;
stus[i].rank=rk;
}

vector<vector<int> > result(M);
vector<int> curRank(M,0);//curRank[i]表示学校i上一个接收学生的排名
for(int i=0;i<N;i++){
for(int j=0;j<K;j++){//检测所有志愿
int apply=stus[i].choice[j];//申请的学校
if(curRank[apply]==stus[i].rank){//与该学校上一个录取的学生同分
school[apply]--;//余量减1
result[apply].push_back(stus[i].id);
break;
}else{
if(school[apply]>0){
school[apply]--;
curRank[apply]=stus[i].rank;
result[apply].push_back(stus[i].id);
break;
}
}
}
}

for(int i=0;i<M;i++){
if(!result[i].empty()){
sort(result[i].begin(),result[i].end());
cout<<result[i][0];
for(auto it=result[i].begin()+1;it!=result[i].end();it++)
cout<<" "<<*it;
}
cout<<endl;
}

return 0;
}

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