PAT_Advanced Level_1080 Graduate Admission(C++_模拟_快排_卡常)

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:

0 10
3
5 6 7
2 8

1 4

Process

emmm，这道题其实就是个快排问题，只要题意读懂了那么很快就能过，但是如果很不幸你也和我一样卡在了第四个测试点，那么下面有几个卡常小技巧，你值得拥有：

• I/O流优化：ios::sync_with_stdio(0)//一旦有了这行函数，代码中就不能出现printf/scanf，只能用cin/cout;
• int前面加register，会快很多;
• 函数前面加inline;

Code

#include<bits/stdc++.h>
using namespace std;
int n, m, k;
struct students {
	double g, ge;
	int rank, id;
	vector<int> want;
};
struct school {
	int num;
	vector<students>schstu;
};
inline bool comp(students a, students b) {
	if (a.g != b.g)
		return a.g > b.g;
	else
		return a.ge > b.ge;
}
inline bool comp2(students a, students b)
{
	return a.id < b.id;
}
int main()
{
	ios::sync_with_stdio(0);
	cin >> n >> m >> k;
	vector<school>sch(m);
	vector<students>stu(n);
	for (register int i = 0; i < m; i++)
		cin >> sch[i].num;
	for (register int i = 0; i < n; i++) {//数据录入
		double g1, g2;
		cin >> g1 >> g2;
		stu[i].id = i;
		stu[i].ge = g1;
		stu[i].g = (g1 + g2) / 2;
		for (register int j = 0; j < k; j++)
		{
			int tempw;
			cin >> tempw;
			stu[i].want.push_back(tempw);
		}
	}
	sort(stu.begin(), stu.end(), comp);//快排
	stu[0].rank = 0;
	for (register int i = 1; i < n; i++) {//排名
		if (stu[i].g == stu[i - 1].g && stu[i].ge == stu[i - 1].ge)
			stu[i].rank = stu[i - 1].rank;
		else
			stu[i].rank = stu[i - 1].rank + 1;
	}
	for (register int i = 0; i < n; i++) {
		students temp = stu[i];
		for (register int j = 0; j < k; j++)
		{
			int wantsch = temp.want[j];//想去的学校
			int num = sch[wantsch].schstu.size();//该校已经录取的学生数量
			if (num < sch[wantsch].num||//未满
				sch[wantsch].schstu[num - 1].rank == temp.rank)//已满但排名相同
			{
				sch[wantsch].schstu.push_back(temp);
				break;
			}
		}
	}
	for (register int i = 0; i < m; i++)
	{
		int num = sch[i].schstu.size();
		if (num)
		{
			sort(sch[i].schstu.begin(), sch[i].schstu.end(), comp2);
			for (int j = 0; j < num; j++)
				cout << sch[i].schstu[j].id << (j != num - 1 ? " " : "\n");
		}
		else
			cout << endl;
	}
	return 0;
}