PAT 1080. Graduate Admission

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

分析

这道题并不是很难，首先按照题意进行排序，后面稍微难处理的点就是当存在排名相同的怎么办，我是用lastaver和lastGe来储存上一位的aver，Ge,，tag数组记录上一位进入的研究院，如果下一位的aver等于lastaver,Ge等于lastGe，说明和上一名同学的排名相同，那么如果他的选择在tag中，那么不管研究院的剩余名额都可以进去。如果排名不相同，则,必须研究院的剩余名额大于0，清空tag然后更新tag的情况。

#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
struct student{
	int id,Ge,Gi,aver;
	int choice[5];
};
bool cmp(const student& s1,const student& s2){
	return s1.aver!=s2.aver?s1.aver>s2.aver:s1.Ge>s2.Ge;
}
int main(){
	int n,m,k;
	cin>>n>>m>>k;
	vector<student> studs(n);
	set<int> addmission[m];
	int num[m],tag[m]={0},lastGe=-1,lastaver=-1;
	for(int i=0;i<m;i++)
	    cin>>num[i];
	for(int i=0;i<n;i++){
		cin>>studs[i].Ge>>studs[i].Gi;
		studs[i].id=i;
		studs[i].aver=(studs[i].Ge+studs[i].Gi)/2.0;
		for(int j=0;j<k;j++)
		    cin>>studs[i].choice[j];
	}
	sort(studs.begin(),studs.end(),cmp);
	for(int i=0;i<n;i++){
		if(studs[i].aver!=lastaver||studs[i].Ge!=lastGe)
			fill(tag,tag+m,0);
		for(int j=0;j<k;j++)
			if(tag[studs[i].choice[j]]==1||num[studs[i].choice[j]]>0){
			    num[studs[i].choice[j]]--;
			   	addmission[studs[i].choice[j]].insert(studs[i].id);
			   	tag[studs[i].choice[j]]=1;
			   	lastaver=studs[i].aver;
			   	lastGe=studs[i].Ge;
			    break;
			}
	}
	for(int i=0;i<m;i++){
			for(auto it=addmission[i].begin();it!=addmission[i].end();it++){
				if(it!=addmission[i].begin())
				   cout<<" ";
				cout<<*it;
			}
		    cout<<endl;
	}
	return 0;
}