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:
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The applicants are ranked according to their final grades, and will be admitted one by one from the top of the rank list.
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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.
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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.
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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题目大意：志愿报考系统。
第一行给定N M K，分别代表学生数量，院校数量，每个学生允许报考的志愿个数
每个学生有个GE分，还有个GI分，学生的最终得分为GE和GI取平均
名次按照总分从高到低排序，如果总分相同，按照GE分排序。如果GE分仍然相同，则这两个人的名次相同
录取时，按照名次从高到低的顺序进行录取，先看第一志愿，再看第二志愿，。。。全部志愿看完后，再看下一个人
如果志愿对应的学校人数满了，但是他的名次与该学校最后一名并列，则仍然可以录取
按照学生的编号顺序(输入时第0-N个)最后输出每个学校录取情况。没有录取到人的学校输出一个换行符即可。
逻辑有些复杂，但是一点点模拟，不会出错的。
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef struct{
int GE, GI, id;
double score; //(GE+GI)/2
vector<int> want;
} stu;
vector<int> limit; //招生总数
vector<vector<int> > ans; //录取信息
vector<stu> V; //学生信息
bool isGay(int a, int b){
return V[a].score == V[b].score && V[a].GE == V[b].GE;
}
int main(){
int N, M, K;
scanf("%d %d %d", &N, &M, &K);
limit.resize(M), ans.resize(M), V.resize(N);
for(int & i : limit) scanf("%d", &i); //读入招生总数
for(int i = 0; i < N; i++){ //读入每个学生的信息
scanf("%d %d", &V[i].GE, &V[i].GI);
V[i].score = 1.0 * (V[i].GE + V[i].GI) / 2;
V[i].want.resize(K);
for(int & j : V[i].want) scanf("%d", &j); //读取志愿
V[i].id = i; //记录学生id
}
sort(V.begin(), V.end(), [](stu a, stu b){ //排序
if(a.score != b.score) return a.score > b.score;
else return a.GE > b.GE;
});
for(int i = 0; i < N; i++){
for(int j : V[i].want){
int this_zhiyuan_ojbk_sum = ans[j].size();
if(this_zhiyuan_ojbk_sum < limit[j] || isGay(i, ans[j][this_zhiyuan_ojbk_sum-1])){
ans[j].push_back(i);
break;
}
}
}
for(auto & it : ans){
sort(it.begin(), it.end(), [](int a, int b){
return V[a].id < V[b].id;
});
for(int i = 0; i < it.size(); i++){
if(i) printf(" ");
printf("%d", V[it[i]].id);
}
printf("\n");
}
return 0;
}