Not Only Algorithm,不仅仅是算法,关注数学、算法、数据结构、程序员笔试面试以及一切涉及计算机编程之美的内容 。。
你的位置:NoAlGo博客 » 题解 » 

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

解答

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int mxsz = 40000 + 10;
struct Stu { int ge, gi, avg, a[6]; } stu[mxsz];
int rnk[mxsz];

bool cmp(const int &a, const int &b)
{
	return stu[a].avg > stu[b].avg || (stu[a].avg == stu[b].avg && stu[a].ge > stu[b].ge); 
}

int main()
{
	int n, m, k, quo[110]; scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < m; i++) scanf("%d", quo+i);
	for (int i = 0; i < n; i++)
	{
		rnk[i] = i;
		scanf("%d%d", &stu[i].ge, &stu[i].gi);
		stu[i].avg = (stu[i].ge + stu[i].gi) / 2;
		for (int j = 0; j < k; j++) scanf("%d", &stu[i].a[j]);
	}
	sort(rnk, rnk+n, cmp); //将下标排序

	vector<int> sch[110];
	for (int i = 0; i < n; i++)
	{
		int id = rnk[i];
		for (int j = 0; j < k; j++)
		{
			int k = stu[id].a[j];
			if (sch[k].size() < quo[k] || (stu[id].avg == stu[sch[k].back()].avg && stu[id].ge == stu[sch[k].back()].ge))
			{
				sch[k].push_back(id);
				break;
			}
		}
	}

	for (int i = 0; i < m; i++)
	{
		sort(sch[i].begin(), sch[i].end());
		for (int j = 0; j < sch[i].size(); j++)
			if (j > 0) printf(" %d", sch[i][j]);
			else printf("%d", sch[i][j]);
		printf("\n");
	}

	return 0;
}
上一篇: 下一篇:

我的博客

NoAlGo头像编程这件小事牵扯到太多的知识,很容易知其然而不知其所以然,但真正了不起的程序员对自己程序的每一个字节都了如指掌,要立足基础理论,努力提升自我的专业修养。

站内搜索

最新评论