标签：PAT 1080 int app 30 grade ge 100 gi

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 G E G_E GE​ , and the terview grade G I G_I GI​​ . The final grade of an applicant is ( G E ​ + G I ​ ) / 2 (G_E​ +G_I​ )/2 (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 G E G_E GE​​ . If still tied, their ranks must be the same.
• Each applicant may have K K 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 ) N (≤40,000) N(≤40,000), the total number of applicants; M ( ≤ 100 ) M (≤100) M(≤100), the total number of graduate schools; and K ( ≤ 5 ) K (≤5) K(≤5), the number of choices an applicant may have.
In the next line, separated by a space, there are M M 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 2+K 2+K integers separated by a space. The first 2 integers are the applicant’s G E G_E GE​​ and G I G_I 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 M−1 M−1, and the applicants are numbered from 0 to N − 1 N−1 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

tips

• 比较 G E + G I G_E+G_I GE​+GI​, 不用除以2, 避免了浮点数比较的问题
• 把所有人排序, 分数高的先选, 分数低的后选, 一旦目标校满了, 如果和目标校的最后一个人分数一样(用vector<pair<int,int> >grade来记录), 那么也可以加入进去, 遍历所有的目标校, 如果被录取了就break出来; 都没录取就落榜gg了
• 部分思路参考了柳神

AC code

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

struct Applicant{
    int ge,gi,id;
    int choice[5];
    bool operator<(const Applicant& b){
        if((ge+gi)!=(b.ge+b.gi)) return (ge+gi)>(b.ge+b.gi);
        if(ge!=b.ge) return ge>b.ge;
        return gi>b.gi;
    }
}app[40005];

int N,M,K;
int quota[105];    //学校的剩余名额

int main(){
    scanf("%d %d %d",&N,&M,&K);
    for(int i=0;i<M;i++){
        scanf("%d",quota+i);
    }
    for(int i=0;i<N;i++){
        app[i].id=i;
        for(int j=0;j<2+K;j++){
            if(j==0) scanf("%d",&app[i].ge);
            else if(j==1) scanf("%d",&app[i].gi);
            else scanf("%d",&app[i].choice[j-2]);
        }
    }
    
    //排名高的先选,败者食尘
    sort(app,app+N);
    vector<vector<int> > admission(M);   //录取名单
    vector<pair<int,int> > grade(M);
     
    for(int i=0;i<N;i++){
        for(int j=0;j<K;j++){
            if(admission[app[i].choice[j]].size()<quota[app[i].choice[j]]){
                admission[app[i].choice[j]].push_back(app[i].id);
                grade[app[i].choice[j]] = make_pair(app[i].ge,app[i].gi);
                break;
            }
            else if(app[i].ge==grade[app[i].choice[j]].first and app[i].gi==grade[app[i].choice[j]].second){
                admission[app[i].choice[j]].push_back(app[i].id);
                break;
            }     
        }
    }
    for(auto& v:admission){
        sort(v.begin(),v.end());
        for(int i=0;i<v.size();i++){
            if(i>0) printf(" ");
            printf("%d",v[i]);
        }
        printf("\n");        
    }
    return 0;
}

标签：PAT,1080,int,app,30,grade,ge,100,gi
来源： https://blog.csdn.net/qq_52429637/article/details/121913138

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。