1080 Graduate Admission (30分)【结构体排序】

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 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

题目大意：

有n位考生，m所学校，每位考生有k个自愿学校，每个学校也有招生人数限制。现在给出所有考生的初试成绩、面试成绩以及k个志愿学校编号，要求模拟学校录取过程，并输出每个学校的录取的考生编号（要求按编号由大到小输出)，下面是录取规则：

1.先按考生的总分从高到低排序，总分相同的按exam grade从高到低排序。如果exam grade仍然相同，则按排名相同处理。

2.按排名先后来考虑每个学生的录取学校。

解题思路：

用一个结构体保存每个考生的相关信息，然后用一个vector数组存储每个录取学生的id。

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;

int quota[110];    //招生名额

struct student {
	int exam_grade;
	int inte_grade;
	int sum;
	int rank;
	int stuID;
	int choices[6];
}stu[40010];

vector<int>admit[110];    //存储学校入取的学生
int last[110];

int cmp(student a, student b)
{
	if (a.sum != b.sum)
		return a.sum > b.sum;
	else return a.exam_grade > b.exam_grade;
}

int main()
{
	int n, m, k;
	cin >> n >> m >> k;
	for (int i = 0; i < m; i++)
	{
		cin >> quota[i];
	}
	for (int i = 0; i < n; i++)
	{
		stu[i].stuID = i;
		cin >> stu[i].exam_grade >> stu[i].inte_grade;
		stu[i].sum = stu[i].exam_grade + stu[i].inte_grade;
		for (int j = 0; j < k; j++)
		{
			cin >> stu[i].choices[j];
		}
	}
	sort(stu, stu + n, cmp);
	for (int i = 0; i < n; i++)
	{
		if (i != 0 && stu[i].sum == stu[i - 1].sum&&stu[i].exam_grade == stu[i - 1].exam_grade)
			stu[i].rank = stu[i - 1].rank;
		else stu[i].rank = i;
	}
	memset(last, -1, sizeof(last));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < k; j++)
		{
			int cho = stu[i].choices[j];    //第i个学生的j个志愿
			int num = admit[cho].size();        //学校招生人数
			int pre = last[cho];
			if (num < quota[cho] || (pre != -1 && stu[i].rank == stu[pre].rank))
			{
				admit[cho].push_back(stu[i].stuID);
				last[cho] = i;
				break;
			}
		}
	}
	for (int i = 0; i < m; i++)
	{
		int len = admit[i].size();
		if (len > 0)
		{
			sort(admit[i].begin(), admit[i].end(), less<int>());
			for (int j = 0; j < admit[i].size(); j++)
			{
				cout << admit[i][j];
				if (j < admit[i].size() - 1)
					cout << " ";
			}
		}
		printf("\n");
	}
	return 0;
}