1142: Graduate Admission(已翻译)
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.
据说,2011年,浙江省约有100所研究生院准备处理4万多份申请。如果你能编写一个程序使录取过程自动化,那就太好了。
每个申请人必须提供两个等级:全国入学考试等级GE和面试等级GI。申请人的最终成绩为(GE + GI) / 2。录取规则如下:
•根据最终成绩进行排名,从排名靠前的依次录取。
•期末成绩持平者,按全国普通入学考试成绩排序。如果成绩仍然是持平,则他们的排名相同。
•每个申请者有K个选择,根据自己的选择进行录取:如果根据排名表,则轮到自己被录取;如果自己心仪学校的名额未满,就会被该学校录取,或者其他的选择也会被依次考虑。如果一个人被所有首选学校拒绝,那么这个不幸的申请人将被拒绝。
•如果排名并列,并且相应的申请人申请的是同一所学校,那么该学校必须录取所有排名相同的申请人,即使超过其名额。
输入格式
Each input file may contain more than 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.
每个输入文件可能包含一个以上的测试用例。
每个案例以包含三个正整数的一行开头:N(≤40000),表示申请总人数;M(≤100),研究生院总数;K(≤5)为申请人可选择的数目。
在下一行,用空格分隔,有M个正整数。第i个整数分别是第i个研究生院的名额。
然后是N行,每行包含2+K个整数,用空格分隔。前两个整数分别是申请人的GE和GI。接下来的K个整数代表首选学校。为了简单起见,我们假设学校的编号从0到M-1,申请人的编号从0到N-1。
输出格式
or 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.
对于每个测试用例,你应该输出所有研究生院的录取结果。每所学校的结果必须占一条线,其中包含学校录取的申请人人数。数字必须按递增顺序排列,并用空格分隔。每行的末尾不能有额外的空格。如果没有申请人被学校录取,则必须输出相应的空行。
输入样例 复制
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
输出样例 复制
0 10
3
5 6 7
2 8
1 4
提示
这一题是解决学校录取考生的问题。只要对考生按一定的顺序排序,然后依次判断考生能否进入自己所选择的学校。
排序是按统考分数GE与面试分数GI的均分非递增进行的,如果均分相同则按统考分数GE非递减进行排序。每个学校都有一定的名额限制,如果有考生与最后一名的排名相同(均分及统考分均相同)则学校需要扩招。
解题步骤如下:
1)读取考生信息,注意记录考生均分及序号;
2)对考生进行排序;
3)对每位考生进行遍历,判断该考生是否能够被自己所选择的学校录取,如果能够录取则对相应的学校添加记录;
4)输出每个学校的信息,按考生的序号非递增输出。