1080. Graduate Admission (30)

1.题目模拟学校录取学生的情况。

2.注意排名,在总分数和各个分数相同的情况下,排名必须相同。

3.注意录取,学校记录一个最后录取学生的排名,当学校没有名额后,判断这个排名和当前申请的学生排名是否相等,如果相等,则必须录取这个学生。

20151124185158330

时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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

AC代码:

//#include<string>
//#include <iomanip>
//#include<stack>
//#include<unordered_set>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<unordered_map>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include <algorithm>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
#include<stack>
using namespace std;

struct StudentNode{
	int Ge, Gi;
	int totalGrade;
	int rank;
	int id;
	vector<int> choice;
	StudentNode() :Ge(0), Gi(0), totalGrade(0), rank(-1), id(-1), choice(0){};
	StudentNode(int x) :Ge(0), Gi(0), totalGrade(0), rank(-1), id(-1), choice(vector<int>(x,0)){};
};
struct SchoolNode{
	int quato;
	int lastRank;
	vector<int> student;
	SchoolNode() :quato(0), lastRank(0), student(0){};
};
bool cmp(const StudentNode&a, const StudentNode&b)
{
	if (a.totalGrade > b.totalGrade)
		return true;
	else if (a.totalGrade == b.totalGrade && a.Ge > b.Ge)
		return true;
	else return false;
}
int main(void)
{
	int studentSum, schoolSum, choiceSum;
	cin >> studentSum >> schoolSum >> choiceSum;
	vector<StudentNode> student(studentSum, StudentNode(choiceSum));
	vector<SchoolNode> school(schoolSum);

	for (int i = 0; i < school.size(); i++)
	{//输入每个学校的招生名额
		scanf("%d", &school[i].quato);
	}

	for (int i = 0; i < student.size(); i++)
	{//输入学生信息
		scanf("%d %d", &student[i].Ge, &student[i].Gi);
		for (int j = 0; j < choiceSum; j++)
		{//输入志愿
			scanf("%d", &student[i].choice[j]);
		}
		student[i].totalGrade = student[i].Ge + student[i].Gi;
		student[i].id = i;
	}

	sort(student.begin(), student.end(), cmp);
	student[0].rank = 0;
	for (int i = 1; i < student.size(); i++)
	{//统计学生的排名
		if (student[i].totalGrade == student[i - 1].totalGrade && student[i].Ge == student[i - 1].Ge)
			student[i].rank = student[i - 1].rank;//所有成绩相等的学生排名一样
		else
			student[i].rank = i;
	}

	for (int i = 0; i < student.size(); i++)
	{
		for (int j = 0; j < student[i].choice.size(); j++)
		{
			int num = student[i].choice[j];
			if (school[num].quato > 0 || (school[num].quato == 0 && school[num].lastRank == student[i].rank))
			{//还有名额,或者,没有名额但是学校招收的最后一名的排名和当前学生排名一样,需要同样招收
				if (school[num].quato>0)
					school[num].quato--;
				school[num].lastRank = student[i].rank;//更新学校招收学生的排名
				school[num].student.push_back(student[i].id);//压入学生的id
				break;//学生被录取了,不再遍历后面的志愿
			}
		}
	}
	for (int i = 0; i < school.size(); i++)
	{
		sort(school[i].student.begin(), school[i].student.end());//对学生的id进行排序
		for (int j = 0; j < school[i].student.size(); j++)
		{
			printf("%d", school[i].student[j]);
			if (j != school[i].student.size() - 1)
				printf(" ");
		}
		printf("\n");
	}

	return 0;
}

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