【题解】【PAT甲】1080 Graduate Admission (30 分)（测试点3）

题目链接

 PTA | 程序设计类实验辅助教学平台

题目描述

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

结尾无空行

Sample Output:

0 10
3
5 6 7
2 8

1 4

结尾无空行

题目大意

 给你M个学院的的录取人数，然后给N个学生，和学生的志愿，按照分高低进行录取，如果总分和考研分一样的话就可以超过名额录取

解题思路

 大部分的数据结构都没什么问题

测试点三是，名额满了，但是有好几个学生的排名相同而且报名不同院校，而且都是超出的那几个人的情况

之前测试点三没过是因为判断名额满了但是排名相同的时候只判断了上一个元素，没有考虑到有多个成绩相同但是报名不同院校的情况

if(ma[v[i].s[j]]==0&&
v[i-1].Gf==v[i].Gf&&
v[i-1].Ge==v[i].Ge&&
v[i-1].choice==v[i].s[j])
//只考虑了上一个人

后面改为用一个数组存储不同学校编号录取最低分的那个人的排名，就可以排除这种情况 

int srank[10];

if(ma[v[i].s[j]]==0&&
v[srank[v[i].s[j]]].Gf==v[i].Gf&&
v[srank[v[i].s[j]]].Ge==v[i].Ge)

题解

#include<bits/stdc++.h>
using namespace std;
struct Node{
	int Ge,Gi;
	int Gf;
	int id;
	int s[8];
	int choice=-1;
};
bool cmp(Node a,Node b){
	return a.Gf==b.Gf?a.Ge>b.Ge:a.Gf>b.Gf;
}
map<int,int> ma;
map<int,vector<int>> school;
int srank[10];
vector<Node> v;
int main(){
	int n,m,k;
	cin>>n>>m>>k;
	for(int i=0;i<m;i++){
		int data;
		scanf("%d",&data);
		ma[i]=data;
	}
	v.resize(n);
	for(int i=0;i<n;i++){
		scanf("%d %d",&v[i].Ge,&v[i].Gi);
		v[i].Gf=v[i].Ge+v[i].Gi;
		v[i].id=i;
		for(int j=0;j<k;j++){
			scanf("%d",&v[i].s[j]);
		}
	}
	sort(v.begin(),v.end(),cmp);
	for(int i=0;i<v.size();i++){
		for(int j=0;j<k;j++){
			//cout<<v[i].s[j];
			//cout<<v[i].id<<"\n";
			if(ma[v[i].s[j]]==0&&
			v[srank[v[i].s[j]]].Gf==v[i].Gf&&
			v[srank[v[i].s[j]]].Ge==v[i].Ge){
				school[v[i].s[j]].push_back(v[i].id);
				v[i].choice=j;
				srank[v[i].s[j]]=i;
				break;
			}
			if(ma[v[i].s[j]]>0){
				school[v[i].s[j]].push_back(v[i].id);
				v[i].choice=v[i].s[j];
				ma[v[i].s[j]]--;
				srank[v[i].s[j]]=i;
				//cout<<v[i].s[j]<<" "<<v[i].id<<" "<<srank[v[i].s[j]]<<'\n';
				break;
			}
		}
	}
	for(int i=0;i<m;i++){
		sort(school[i].begin(),school[i].end());
		for(int j=0;j<school[i].size();j++){
			if(j!=0)	cout<<" ";
			cout<<school[i][j];
		}
		cout<<endl;
	}
}