PAT甲级练习1012. The Best Rank (25)

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本文介绍了一个学生排名系统的实现方法，该系统能够根据学生的三门课程成绩（C语言编程、数学、英语）及其平均分来确定每位学生的最佳排名，并按最优排名进行展示。

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1012. The Best Rank (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90
310102     70 95 88 84
310103     82 87 94 88
310104     91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

Sample Input

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output

1 C
1 M
1 E
1 A
3 A
N/A

这题其实还算简单吧，主要碰到的就是超时的问题。。刚开始是建了一个内部类存储每个学生的相关数据，果断超时。后来就直接运用数组来做了，但不知道怎么优化比较好，运行结果也比较悬。。

import java.util.Scanner;

public class Main {

	static String[] subjects = {"A", "C", "M", "E"};
	static int[] IDs = new int[10000000];
	static int[] checkIDs = new int[2001];
	static int[][] grades = new int[2001][4];
	
	public static void main(String[] args) {		
			
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();//the total number of students
		int m = s.nextInt();//the number of the checks
		
		for(int i=1; i<=n; i++){
			IDs[s.nextInt()] = i;
			grades[i][1] = s.nextInt();
			grades[i][2] = s.nextInt();
			grades[i][3] = s.nextInt();
			grades[i][0] = (grades[i][1] + grades[i][2] + grades[i][3])/3;
		}
		
		for(int i=0; i<m; i++){
			checkIDs[i] = s.nextInt();
		}
		
		for(int i=0; i<m; i++){
			int[] ranks = {1,1,1,1};
			int exist = IDs[checkIDs[i]];
			if(exist == 0){
				System.out.printf("N/A\n");
			}else{
				for(int j=1; j<=n; j++){
					for(int k=0; k<4;k++){
						if(grades[j][k]>grades[exist][k]) ranks[k]++;
					}
				}
				int x=0, top=2001;
				for(int k=0;k<4;k++){
					if(ranks[k]<top){
						top = ranks[k];
						x = k;
					}
				}
				System.out.printf("%d %s\n", top, subjects[x]);
			}
		}
	}
}