PAT-1080 Graduate Admission

1080 Graduate Admission (30)（30 分）

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 G~E~, and the interview grade G~I~. The final grade of an applicant is (G~E~ + G~I~) / 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~. 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 G~E~ and G~I~, 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

 根据成绩排序，从高分开始录取，注意相同排名的学生可以超额录取。学校录取最低分可以判断是否需要超额录取。之前有一个测试点一直超时，以为是STL容器搞多了，尝试了多种写法，其实不是，STL对时间的消耗很少，嵌套for循环（即使感觉每个循环只循环几次）的影响要大几个数量级，以后优化时间要从减少循环下手。

#include<stdio.h>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

typedef struct stu{
	int id,ge,gi;
	int ch[5];
}stu;
int n,m,K;
int quota[105];
int lowestge[105];
int lowestgi[105];
priority_queue<int,vector<int>,greater<int> > result[105];//录取结果
vector<stu> s;

bool cmp(stu s1,stu s2){
	if((s1.ge+s1.gi)!=(s2.ge+s2.gi))
		return (s1.ge+s1.gi)>(s2.ge+s2.gi);
	else{
		if(s1.ge!=s2.ge)
			return s1.ge>s2.ge;
		else return false;
	}
}
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++){
		stu st;st.id=i;
		scanf("%d %d",&st.ge,&st.gi);
		for(int k=0;k<K;k++){
			scanf("%d",&st.ch[k]);
		}
		s.push_back(st);
	}
	sort(s.begin(),s.end(),cmp);
	fill(lowestge,lowestge+m,-1);
	fill(lowestgi,lowestgi+m,-1);

	for(int i=0;i<s.size();i++){
		for(int j=0;j<K;j++){
			int sch=s[i].ch[j];
			if(quota[sch]>0||
				(lowestgi[sch]==s[i].gi&&lowestge[sch]==s[i].ge)){
					result[sch].push(s[i].id);		
					lowestgi[sch]=s[i].gi;
					lowestge[sch]=s[i].ge;
					quota[sch]--;
					break;
			}
		}
	}

	for(int i=0;i<m;i++){
		if(result[i].size()==0){
			printf("\n");
			continue;
		}
		printf("%d",result[i].top());result[i].pop();
		for(;!result[i].empty();result[i].pop())
			printf(" %d",result[i].top());

		printf("\n");
	}

	return 0;
}