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
Sample Output:
1 C
1 M
1 E
1 A
3 A
N/A
题意:
n个学生，现分别给出每个学生的3科成绩C、M、E，又有A为平均分
给出m个查询，输入学生ID，输出A、C、M、E中最高的排名及对应的是哪个
①若多个排名相同，则按A>C>M>E输出最高优先级的
②若查询的学生ID不存在，则输出N/A
思路：
0、对比分析
【A1012】在《算法笔记（胡凡）》中，按训练顺序是排在【A1025】和【A1062】两道排序题之后
因为那两题是最基础的仅单项数据的排序，而这一题扩展到了多项（4项）数据的排序
在刚看到这题时因为要进行多项数据的排序而陷入了混乱
现在复盘梳理时，我认为 从 单项数据排序 扩展到 多项数据排序 的思路会更加清晰且易于理解
1、先从单项数据排序开始
我们假设现在只有一门科目X
①结构体
1 struct Student {
2 int id; //学生id
3 int grade; //分数
4 }stu[2010]; //题目给出N《=2000
②排序函数cmp
1 bool cmp(Student a, Student b) {
2 return a.grade> b.grade;
3 }
③存放排名的数组Rank（※这里踩过一个坑：rank是关键字，所以要写作Rank)
在之前的【A1025】和【A1062】中，id只用于输出，故都是常规地使用char类型数组存放
而这一题中，id要担当从 分数Student.grade 到 排名Rank 的对应 的责任
如果在Student中将id设为char[6]，那Rank中就既要存放int型的排名，又要存放char[]型的id——显然，就算强行实现，也非常麻烦
考虑到题目中的id为6位0-9的数字，故我们将id设为int类型，而Rank的大小则开为[1000000]，其下标就包含了100000~999999的所有可能的ID
即Rank[Student.id] = Student.grade，每个下标对应的即为此ID学生的排名
1 int Rank[10000000] = {0};
④main()
1 int main() {
2 //总计有n个学生，m个查询
3 scanf("%d%d", &n, &m);
4 for (int i = 0; i < n; i++) {
5 scanf("%d%d", &stu[i].id, &stu[i].grade);
6 }
7 //对n个学生的成绩进行排序
8 sort(stu, stu + n, cmp);
9 //第一名的排名为1
10 Rank[stu[0].id] = 1;
11 //剩下学生的排名
12 for (int i = 1; i < n; i++) {
13 if (stu[i].grade == stu[i - 1].grade) {
14 Rank[stu[i].id] = Rank[stu[i - 1].id];
15 } else {
16 Rank[stu[i].id] = i + 1;
17 }
18 }
19 //m个查询
20 for (int i = 0; i < m; i++) {
21 scanf("%d", &query);
22 if (Rank[query] == 0) {
23 printf("N/A\n");
24 } else {
25 printf("%d %c\n", Rank[query]);
26 }
27 }
28 return 0;
29 }
2、再扩展到多项数据排序
①结构体修改
1 struct Student {
2 int id;
3 int grade[4];
4 }stu[2010];
②排序函数cmp修改
1 int now; //标识当前是A、C、M、E中的哪个科目
2
3 bool cmp(Student a, Student b) {
4 return a.grade[now] > b.grade[now];
5 }
③还需要的两个数组
1 char course[4] = {‘A‘, ‘C‘, ‘M‘, ‘E‘}; //按优先级排序更方便
2
3 int Rank[10000000][4] = {0};
④main() : 录入成绩
1 for (int i = 0; i < n; i++) {
2 scanf("%d%d%d%d", &stu[i].id, &stu[i].grade[1], &stu[i].grade[2], &stu[i].grade[3]); //C、M、E
3 stu[i].grade[0] = round((stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3]) / 3.0) + 0.5; //A
4 }
⑤main() : 排名
1 for (now = 0; now < 4; now++) { //四个科目分别排
2
3 //内部就和单数据排序相同了
4 sort(stu, stu + n, cmp);
5 Rank[stu[0].id][now] = 1;
6 for (int i = 1; i < n; i++) {
7 if (stu[i].grade[now] == stu[i - 1].grade[now]) {
8 Rank[stu[i].id][now] = Rank[stu[i - 1].id][now];
9 } else {
10 Rank[stu[i].id][now] = i + 1;
11 }
12 }
13
14 }
⑥main() : 查询
1 for (int i = 0; i < m; i++) {
2 scanf("%d", &query);
3 if (Rank[query][0] == 0) {
4 printf("N/A\n");
5 } else {
6 int k = 0; //因有A>C>M>E的优先级，故这里设初值0
7 for (int j = 0; j < 4; j++) {
8 if (Rank[query][j] < Rank[query][k]) {
9 k = j;
10 }
11 }
12 printf("%d %c\n", Rank[query][k], course[k]);
13 }
14 }
3、题解
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #include<algorithm>
5 using namespace std;
6 struct Student {
7 int id;
8 int grade[4];
9 }stu[2010];
10
11 char course[4] = {‘A‘, ‘C‘, ‘M‘, ‘E‘};
12 int Rank[10000000][4] = {0};
13 int now;
14
15 bool cmp(Student a, Student b) {
16 return a.grade[now] > b.grade[now];
17 }
18 int main() {
19 int n, m, query;
20 scanf("%d%d", &n, &m);
21 for (int i = 0; i < n; i++) {
22 scanf("%d%d%d%d", &stu[i].id, &stu[i].grade[1], &stu[i].grade[2], &stu[i].grade[3]);
23 stu[i].grade[0] = round((stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3]) / 3.0) + 0.5;
24 }
25 for (now = 0; now < 4; now++) {
26 sort(stu, stu + n, cmp);
27 Rank[stu[0].id][now] = 1;
28 for (int i = 1; i < n; i++) {
29 if (stu[i].grade[now] == stu[i - 1].grade[now]) {
30 Rank[stu[i].id][now] = Rank[stu[i - 1].id][now];
31 } else {
32 Rank[stu[i].id][now] = i + 1;
33 }
34 }
35 }
36 for (int i = 0; i < m; i++) {
37 scanf("%d", &query);
38 if (Rank[query][0] == 0) {
39 printf("N/A\n");
40 } else {
41 int k = 0;
42 for (int j = 0; j < 4; j++) {
43 if (Rank[query][j] < Rank[query][k]) {
44 k = j;
45 }
46 }
47 printf("%d %c\n", Rank[query][k], course[k]);
48 }
49 }
50 return 0;
51 }
原文地址：https://www.cnblogs.com/Bananice/p/12238691.html