Graduate Admission (30)

Graduate Admission (30)

难度 ： ⭐⭐⭐
题目连接
题目描述
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.

输入描述:
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.

输出描述:
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.

输入例子:
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

输出例子:
0 10
3
5 6 7
2 8

1 4

大意
给出学生成绩，填报的志愿，还有学校招生人数等数据，根据高考录取考试的规则，求出每个学校最后录取到的学生并输出。
录取规则：1.学生有两门科目，先按平均分做比较依据，平均分相等时用第一门科目做依据，若都相等则排名并列； 2.当有多个排名相同的学生选择了同一所学校时，学校应该录取这些学生，即是会操场预计招生数；3…

分析
模拟题，难度主要集中在阅读和理解题意。清楚了情景和要求就不难做了。

MYCODE

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#define sum(i) i.s1 + i.s2
using namespace std;

struct School {
	int minRank;	//录取的最低排名
	int stuNum;		//计划收取的学生数
	vector<int> stu;	//录取的学生
};

struct Student{
	int index;		//学生编号
	int s1, s2;		//科目一和科目二分数
	int rank;		//排名
	int choose[6];	//志愿
};

int stuNum, schNum, choNum;

bool func(Student &a, Student &b){
	if (sum(a) != sum(b)) return sum(a) > sum(b);
	return a.s1 > b.s1;
}

School school[101];
Student student[40001];

int main(){
	scanf("%d%d%d", &stuNum, &schNum, &choNum);
	for (int i = 0; i < schNum; i++) scanf("%d", &school[i].stuNum);
	for (int i = 0; i < stuNum; i++){
		student[i].index = i;
		scanf("%d%d", &student[i].s1, &student[i].s2);
		for (int j = 0; j < choNum; j++) scanf("%d", &student[i].choose[j]);
	}
	sort(student, student + stuNum, func);
	student[0].rank = 1;
	//为学生成绩排名
	for (int i = 1; i < stuNum; i++) {
		if (sum(student[i]) == sum(student[i - 1]) && student[i].s1 == student[i - 1].s1) {
			student[i].rank = student[i - 1].rank;
		}
		else{
			student[i].rank = student[i - 1].rank + 1;
		}
		//cout << "------->" <<i << "   " << student[i].rank << endl;
	}
	//学生选择学校
	for (int i = 0; i < stuNum; i++){
		for (int j = 0; j < choNum; j++) {
			int want = student[i].choose[j];
			if (school[want].stuNum > 0 || school[want].minRank == student[i].rank){
				school[want].stu.push_back(student[i].index);
				school[want].stuNum--;
				school[want].minRank = student[i].rank;
				break;
			}
		}
	}
	//输出录取状况
	for (int i = 0; i < schNum; i++){
		sort(school[i].stu.begin(), school[i].stu.end());
		if (school[i].stu.empty()) {
			cout << endl;
			continue;
		}
		cout << school[i].stu[0];
		for (int j = 1; j < school[i].stu.size(); j++) cout <<" "<<school[i].stu[j] ;
		cout << endl;
	}
}