PAT A 1080. Graduate Admission (30)

题目

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 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

 

直接模拟即可，注意点：

1、最终分数是带有小数的，可以直接求和两个分数代表最终分数

2、注意录取规则的最后一条

 

代码：

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

const int MAX=0x3fffffff;

struct stu	//学生结构
{
	int id;	//id
	int GE,G;	//GE分，总分（只是总和）
	int choice[5];	//志愿
};

bool cm(const stu &s1,const stu &s2);	//根据总分,GE分排序

int main()
{
	int n,m,k;
	cin>>n>>m>>k;

	int num[100];	//每个学校人数限制
	stu* student=new stu[n];	//所有的学生
	
	int i,j;
	for(i=0;i<m;i++)	//输入信息
		scanf("%d",&num[i]);
	int GI;
	for(i=0;i<n;i++)
	{
		student[i].id=i;
		scanf("%d %d",&student[i].GE,&GI);
		student[i].G=student[i].GE+GI;	//总分不必除2，GI分不用保留
		for(j=0;j<k;j++)
			scanf("%d",&student[i].choice[j]);
	}
	sort(student,student+n,cm);	//排序

	int num_ac[100]={0};	//到某个名次时各个学校已经录取的人数
	vector<int> school[100];	//保留各个学校录取的学生id
	i=0;j=0;
	int p,q;
	while(i<n)	//i为并列名次段的第一个，j为超尾
	{
		while(student[j].G==student[i].G&&student[j].GE==student[i].GE)	//寻找超尾
			j++;
		for(p=i;p<j;p++)	//扫描名次段学生
		{
			for(q=0;q<k;q++)	//处理志愿
			{
				if(num_ac[student[p].choice[q]]<num[student[p].choice[q]])
				{
					school[student[p].choice[q]].push_back(student[p].id);
					break;
				}
			}
		}
		for(p=0;p<m;p++)	//一个名次段结束后更新录取人数
			num_ac[p]=school[p].size();
		i=j;
	}

	for(i=0;i<m;i++)	//对各个学校的录取人数排序，输出
	{
		if(school[i].empty())
		{
			printf("\n");
			continue;
		}
		sort(school[i].begin(),school[i].end());
		printf("%d",school[i][0]);
		for(j=1;j<school[i].size();j++)
			printf(" %d",school[i][j]);
		printf("\n");
	}
	
	delete [] student;

	return 0;
}

bool cm(const stu &s1,const stu &s2)
{
	if(s1.G>s2.G)
		return true;
	else if(s1.G==s2.G)
		return s1.GE>s2.GE;
	else
		return false;
}