pat 1080. Graduate Admission(30)浙大再試験上機第四題
1080. Graduate Admission (30)
時間の制限
200 ms
メモリ制限
32000 kB
コード長制限
16000 B
クイズルーチン
Standard
作成者
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:
Sample Output:
コミットコード
考え方:
1:学生は指定した方法でソートし、ソートの優先度は最終点数が高い->geが高い->で、どちらも同じで、等しい.
2:そして、順番を決めた学生の順位に基づいて、学生ごとにその順位を指定し、この順位が同じ等級の順位であることに注意してください.例えば(90,90)、(90,90)、(89,91)、1,2です.
3:学生にランキングを指定するのは、主に学校に対して、もし一人の学生がちょうどこの学校に入ったら、他のこの学生と同じ順位の学生は、この学校を選んだら、入る資格があるからです.
学校の学生募集人数の影響を受けない.
4:それから、高いところから低いところまで、学生の志望を一度にチェックします.学生が学校を申請するたびに、その順位を学校に教えなければならない.このように、学校は学生募集がいっぱいになったとき、ランキングとその最低を受け取ると
学生募集の順位が同じ学生は、この学生を採用することができます.
5:次に、各学校で採用された学生のidを印刷し、低から高出力に並べ替える.余計な情報を出力しないでください.
時間の制限
200 ms
メモリ制限
32000 kB
コード長制限
16000 B
クイズルーチン
Standard
作成者
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
コミットコード
考え方:
1:学生は指定した方法でソートし、ソートの優先度は最終点数が高い->geが高い->で、どちらも同じで、等しい.
2:そして、順番を決めた学生の順位に基づいて、学生ごとにその順位を指定し、この順位が同じ等級の順位であることに注意してください.例えば(90,90)、(90,90)、(89,91)、1,2です.
3:学生にランキングを指定するのは、主に学校に対して、もし一人の学生がちょうどこの学校に入ったら、他のこの学生と同じ順位の学生は、この学校を選んだら、入る資格があるからです.
学校の学生募集人数の影響を受けない.
4:それから、高いところから低いところまで、学生の志望を一度にチェックします.学生が学校を申請するたびに、その順位を学校に教えなければならない.このように、学校は学生募集がいっぱいになったとき、ランキングとその最低を受け取ると
学生募集の順位が同じ学生は、この学生を採用することができます.
5:次に、各学校で採用された学生のidを印刷し、低から高出力に並べ替える.余計な情報を出力しないでください.
//15:17 reading-> 15:23 thinking ->15:28 coding->16:20 finished
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct Students
{
int id;
double g1;
double g2;
double fg;
int rank;
int apps[7];
};
struct School
{
int quota;
vector<int> stus;
int lowerRank;
};
Students stus[40003];
School schs[103];
int cmp(const void *a,const void *b)
{
Students x = *((Students *)a);
Students y = *((Students *)b);
if(x.fg!=y.fg)
{
return (x.fg>y.fg)?-1:1;
}else if(x.g1!=y.g1)
{
return (x.g1>y.g1)?-1:1;
}else
{
return 0;
}
}
int main()
{
int n,m,k;
int i,j,qu,s;
scanf("%d%d%d",&n,&m,&k);
for(i=0;i<m;i++)
{
scanf("%d",&schs[i].quota);
schs[i].lowerRank = -1;
}
for(i=0;i<n;i++)
{
scanf("%lf%lf",&stus[i].g1,&stus[i].g2);
stus[i].fg = (stus[i].g1+stus[i].g2)/2;
stus[i].id = i;
for(j=0;j<k;j++)
{
scanf("%d",&s);
stus[i].apps[j] = s;
}
}
qsort(&stus[0],n,sizeof(Students),cmp);
int r=1,same = 0;
i = 0;
while(true)
{
same = 0;
if(i>=n)
{
break;
}
stus[i].rank = r;
while(i<n-1&&stus[i].fg==stus[i+1].fg &&
stus[i].g1==stus[i+1].g1)
{
stus[i+1].rank = r;
same++;
i++;
}
r += same;
r++;
i++;
}
for(i=0;i<n;i++)
{
for(j=0;j<k;j++)
{
s = stus[i].apps[j];
if(schs[s].stus.size()<schs[s].quota)
{
schs[s].stus.push_back(stus[i].id);
schs[s].lowerRank = stus[i].rank;
break;
}else
{
if(schs[s].lowerRank==stus[i].rank)
{
schs[s].stus.push_back(stus[i].id);
break;
}
}
}
}
for(i=0;i<m;i++)
{
bool flag = false;
if(schs[i].stus.size()>0)
{
//printf("%d ",i);
sort(schs[i].stus.begin(),schs[i].stus.end());
for(j=0;j<schs[i].stus.size();j++)
{
printf("%d",schs[i].stus[j]);
if(j!=schs[i].stus.size()-1)
{
printf(" ");
}
}
}else
{
printf("
");
flag = true;
}
if(i!=m-1&&!flag)
{
printf("
");
}
}
return 0;
}