PAT甲级 1080 Graduate Admission

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 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

这道题虽然是30分的题目，但是真的没有什么高深莫测的算法，按部就班，sort大法好，注意题目要求看清读懂。

/**
* Copyright(c)
* All rights reserved.
* Author : csdn写完这道题就睡觉
* Description : JZU PAT test Level: A 
*/
#include 
#include 
#include 
using namespace std;
int N,M,K;
int quota[101]; //院校研究生录取名额
struct applicant{
	int ID;//申请者序号
	int Ge,Gi;
	int grade; //总成绩 不知道 20 和 21 是否有区别 /2=10 
	int choice[6]; 
}; 
vector<applicant> apps; //申请信息 
vector<applicant> admission[101];//录取信息 
bool cmp(applicant a,applicant b)
{
 	if(a.grade!=b.grade)
		return a.grade>b.grade;
	return a.Ge>b.Ge;	
}
bool cmp2(applicant a,applicant b)
{
	return a.ID<b.ID;
}
int main()
{
	cin>>N>>M>>K;
	int i,j;
	applicant t;
	for(i=0;i<M;i++)
		scanf("%d",&quota[i]);
	for(i=0;i<N;i++)
	{
		t.ID=i;
		scanf("%d %d",&t.Ge,&t.Gi);
		t.grade=t.Ge+t.Gi;
		for(j=0;j<K;j++)
			scanf("%d",&t.choice[j]); //报考院校顺序	
		apps.push_back(t);	//所有学生信息放入apps中		 
	} 
  	//决定学生是否录取从学生角度出发而不是学校角度
	sort(apps.begin(),apps.end(),cmp); //学生成绩排名 
	for(i=0;i<N;i++) //对N个学生按成绩由高到低逐一录取 
	{
		for(j=0;j<K;j++) //志愿顺序录取 
		{
			int c=apps[i].choice[j]; //当前学生的志愿学校 
			if(quota[c]>0)//如果院校c还有名额，
			{
				quota[c]--;
				admission[c].push_back(apps[i]);
				break;
			}
			else if(!admission[c].empty())
			{  		
  			  	t=admission[c].back(); //t为学校c当前录取的最后一名 
			   if(apps[i].grade==t.grade && apps[i].Ge==t.Ge )//同样的名次 
			   {
   				admission[c].push_back(apps[i]);
				break;
			   }	 	
			}  
		}
	}
	for(i=0;i<M;i++)
	{
		sort(admission[i].begin(),admission[i].end(),cmp2);
		for(j=0;j<admission[i].size();j++)
		{
			printf("%d",admission[i][j].ID);
			if(j!=admission[i].size()-1)
			printf(" ");
		}
		printf("\n");
	}
	return 0;
}