(PAT)The Best Rank(哈希查找加排序)

川长思鸟来 2022-03-28 04:19 151阅读 0赞

1012 The Best Rank （25 分）

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 Algrbra), 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 Specification: ###

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 Specification: ###

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

###  ###

具体思路:

使用哈希表存储学生4门成绩的排名就可以了:

int srank\[1000000\]\[4\]; //一维数组代表学号,二维记录各门课程的排名

读入学生的成绩和数据后,利用哈希表记录每门课的排名,排名记录算法如下:

int re = 1;
    		for (int i = 0; i < N; ++i) {
    			if (i > 0 && Students[i].grades[now] != Students[i - 1].grades[now]) {
    				re = i + 1;
    			}
    			srank[Students[i].id][now] = re;
    		}

查询的话利用哈希表查找即可

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    
    struct Student {
    	int id;
    	int grades[4];
    }Students[2010];
    //新技能,利用哈希表进行查询
    int srank[1000000][4];   //哈希表
    char Course[4] = { 'A','C','M','E' };
    int now;
    bool cmp(Student a, Student b) {
    	return a.grades[now] > b.grades[now];
    }
    int queres[2010];
    int main() {
    	memset(srank, 0, sizeof(srank));
    	int N, M;
    	scanf("%d %d", &N, &M);
    	for (int i = 0; i < N; ++i) {
    		scanf("%d %d %d %d", &Students[i].id, &Students[i].grades[1], &Students[i].grades[2], &Students[i].grades[3]);
    		Students[i].grades[0] = round((Students[i].grades[1] + Students[i].grades[2] + Students[i].grades[3]) / 3.0)+0.5;
    	}
    
    	for (now = 0; now < 4; ++now) {
    		sort(Students, Students + N, cmp);
    		int re = 1;
    		for (int i = 0; i < N; ++i) {
    			if (i > 0 && Students[i].grades[now] != Students[i - 1].grades[now]) {
    				re = i + 1;
    			}
    			srank[Students[i].id][now] = re;
    		}
    	}
    	//查询操作
    	for (int i = 0; i < M; ++i) {
    		scanf("%d", &queres[i]);
    	}
    	for (int i = 0; i < M; ++i) {
    		if (srank[queres[i]][0] == 0) {
    			printf("N/A\n");
    			continue;
    		}
    
    		int k = 0;
    		for (int j = 0; j < 4; ++j) {
    			if (srank[queres[i]][j] <srank[queres[i]][k]) {
    				k = j;
    			}
    		}
    		printf("%d %c\n", srank[queres[i]][k], Course[k]);
    	}
    	system("PAUSE");
    	return 0;
    }