1080 Graduate Admission (30 分) PAT 甲级模拟考研我的代码+别人的代码

悠悠 2024-02-19 21:35 50阅读 0赞

1080 Graduate Admission (30 分)

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

第一次做30分的题，

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<map>
    #include<algorithm>
    using namespace std;
    struct student{
    	int ge,gi;
    	double zg;
    	int choices[6];
    	bool flag;
    	int id;
    };
    struct school{
    	int num;
    	int s[40000];
    	int last;
    	int l;
    };
    student a[40002];
    school b[101];
    void init(int n,int m,int k){
    	for(int i=0;i<n;i++)
    	{
    		a[i].ge=0;
    		a[i].gi=0;
    		a[i].zg=0;
    		fill(a[i].choices,a[i].choices+k,-1);
    		a[i].flag=false;
    		a[i].id=-1;
    	}
    	for(int i=0;i<m;i++)
    	{
    		b[i].num=0;
    		fill(b[i].s,b[i].s+k,-1);
    		b[i].last=0;
    		b[i].l=0;
    	}
    }
    int cmp(student a,student b)
    {
    	if(a.zg!=b.zg)
    	return a.zg>b.zg;
    	else
    	if(a.ge==b.ge)
    	return a.id<b.id;
    	else
    	return a.ge>b.ge;
    }
    int main()
    {
    	map<int,student> c;
    	c.clear();
    	int n,m,k,i,j,t1,t2,t3,len;
    	cin>>n>>m>>k;
    	init(n,m,k);
    	for(i=0;i<m;i++)
    	cin>>b[i].num;
    	for(i=0;i<n;i++)
    	{
    		cin>>t1>>t2;
    		a[i].ge=t1;
    		a[i].gi=t2;
    		a[i].zg=(a[i].gi+a[i].ge)/2.0;
    		a[i].id=i;
    		for(j=0;j<k;j++)
    		{
    			cin>>t3;
    			a[i].choices[j]=t3;
    		}
    	}
    	sort(a,a+n,cmp);
    	for(i=0;i<n;i++)
    	c.insert(make_pair(a[i].id,a[i]));
    	len=n;
    	for(i=0;i<len;i++)
    	{
    		for(j=0;j<k;j++)
    		{
    		    int e=a[i].choices[j];   // 这个地方是重点。
    			if(b[a[i].choices[j]].num>0||(b[e].last-1>=0&&b[e].l!=0&&   /c[b[e].s[b[e].last-1]].zg-a[i].zg<=0.0001&&c[b[e].s[b[e].last-1]].ge==a[i].ge))
    			{
    				b[e].s[b[e].last++]=a[i].id;
    				b[e].num--;
    				b[e].l++; 
    				a[i].flag=true;
    				break;
    			}
    		}
    	}
    	for(i=0;i<m;i++)
    	{
    		sort(b[i].s,b[i].s+b[i].l);
    		len=b[i].l;
    		for(j=0;j<len;j++)
    		{
    			if(j==0)
    			printf("%d",b[i].s[j]);
    			else
    			printf(" %d",b[i].s[j]);
    		}
    		printf("\n");
    	}
    	return 0;
    }

别人的代码：

#include<cstdio>
    #include<algorithm>
    using namespace std;
    struct student
    {
    	int id;
    	int ge, gi;//初试和复试成绩
    	int total;//总成绩
    	int a[6];//志愿学校数
    	int rank;//总的排名
    }stu[40010];
    struct school
    {
    	int luqu[40010];//录取人的id
    	int peier;//每个学校的配额人数
    	int sum=0;//实际录取数
    	int last=-1;//最后录取人的id,初始为-1，表示没有人被录取
    }sch[110];
    bool cmp(student a, student b)
    {
    	if (a.total != b.total) return a.total > b.total;
    	else  return a.ge > b.ge;//注意这里给出else
    }
    bool cmpid(int a, int b)//学校输出录取情况按照学生id
    {
    	return stu[a].id < stu[b].id;
    }
    int main()
    {
    	int n, m, k;//n-总人数 m-总学校数 k-每个人的志愿数
    	scanf("%d %d %d", &n, &m, &k);
    	for (int i = 0; i < m; i++)
    	{
    		scanf("%d", &sch[i].peier);
    	}
    	for (int i = 0; i < n; i++)
    	{
    		scanf("%d %d", &stu[i].ge, &stu[i].gi);
    		stu[i].id = i;
    		stu[i].total = stu[i].ge + stu[i].gi;
    		for (int j = 0; j < k; j++)
    		{
    			scanf("%d", &stu[i].a[j]);
    		}
    	}
    	sort(stu, stu + n, cmp);//按照cmp排序
    
    
    	//接下来开始按照分数由高到低，从上至下确定每一个学生的录取情况
    	//1、处理rank
    	stu[0].rank = 1;
    	for (int i = 1; i < n; i++)
    	{
    		if (stu[i].total == stu[i - 1].total&&stu[i].ge == stu[i - 1].ge) stu[i].rank = stu[i - 1].rank;
    		else stu[i].rank = i + 1;
    	}
    
    
    	//2、录取
    	for (int i = 0; i < n; i++)
    	{
    		for (int j = 0; j < k; j++)
    		{
    			if (sch[stu[i].a[j]].peier != 0 || stu[sch[stu[i].a[j]].last].rank == stu[i].rank)
    			//当志愿学校的配额不为0或者自己名次与最后录取人相同
    			{
    				//一开始把下面的j全写成k了，然后看了半天没看出来，被自己的zz深深打动了
    				sch[stu[i].a[j]].luqu[sch[stu[i].a[j]].sum] = i;//记录录取人的i 而不是id
    				sch[stu[i].a[j]].last = i;//记录last
    				sch[stu[i].a[j]].sum++;//录取总人数加一
    				if (sch[stu[i].a[j]].peier != 0) sch[stu[i].a[j]].peier--;//配额不为0，配额减一
    				break;
    			}
    		}
    	}
    
    
    
    	//输出每个学校的录取情况
    	for (int i = 0; i < m; i++)
    	{
    		if (sch[i].sum != 0)
    		{
    			sort(sch[i].luqu, sch[i].luqu + sch[i].sum, cmpid);//这个sort自己用的不熟练
    			for (int j = 0; j < sch[i].sum; j++)
    			{
    				printf("%d", stu[sch[i].luqu[j]].id);//输出时候注意输出的是id
    				if (j != sch[i].sum-1) printf(" ");
    			}
    		}
    			printf("\n");
    	}
    	
    	return 0;
    }