PAT——甲级1012：The Best Rank（有坑）

1012 The Best Rank （25 point(s)）

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

这个题倒是不难，但是挺复杂的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Student
{
    char id[];
    int A, C, M, E;//每个同学的分数
    int A_rank, C_rank, M_rank, E_rank;//每个分数的排名
}stu[];
bool cmp_A(Student a, Student b) { return a.A > b.A; }//这四个函数是排序的时候用的，分别对四个成绩排序。
bool cmp_C(Student a, Student b) { return a.C > b.C; }
bool cmp_M(Student a, Student b) { return a.M > b.M; }
bool cmp_E(Student a, Student b) { return a.E > b.E; }
void print_max(int a, int c, int m, int e) {//这个函数打印出来最高排名。a c m e 分别代表的是各成绩排名。
    if (a <= c && a <= m && a <= e)printf("%d A\n", a);//注意这里的=，符合优先级A>C>M>E
    else if (c < a&&c <= m && c <= e)printf("%d C\n", c);
    else if (m < a&&m < c &&  m <= e)printf("%d M\n", m);
    else printf("%d E\n", e);
}
int main() {
    int m, n;
    scanf("%d%d", &n, &m);
    for (int i = ;i < n;i++) {
        scanf("%s%d%d%d", stu[i].id, &stu[i].C,&stu[i].M,&stu[i].E);
        stu[i].A = (stu[i].C+stu[i].M + stu[i].E) / ;
    }

    sort(stu, stu + n, cmp_A);//对A成绩排序
    stu[].A_rank = ;
    for (int i = ;i < n;i++) { //对A成绩排名
        if (stu[i].A == stu[i - ].A)
            stu[i].A_rank = stu[i - ].A_rank;
        else
            stu[i].A_rank = i + ;
    }
    //对C成绩排序，排名
    sort(stu, stu + n, cmp_C);
    stu[].C_rank = ;
    for (int i = ;i < n;i++) {
        if (stu[i].C == stu[i - ].C)
            stu[i].C_rank = stu[i - ].C_rank;
        else
            stu[i].C_rank = i + ;
    }
    //对M成绩排序，排名
    sort(stu, stu + n, cmp_M);
    for (int i = ;i < n;i++)  stu[i].M_rank = i + ;
    stu[].M_rank = ;
    for (int i = ;i < n;i++) {
        if (stu[i].M == stu[i - ].M)
            stu[i].M_rank = stu[i - ].M_rank;
        else
            stu[i].M_rank = i + ;
    }
    //对E成绩排序，排名
    sort(stu, stu + n, cmp_E);
    for (int i = ;i < n;i++)  stu[i].E_rank = i + ;
    stu[].E_rank = ;
    for (int i = ;i < n;i++) {
        if (stu[i].E == stu[i - ].E)
            stu[i].E_rank = stu[i - ].E_rank;
        else
            stu[i].E_rank = i + ;
    }
    for (int i=;i < m;i++) {
        char M_ID[];
        bool flag = false;
        scanf("%s", M_ID);
        for (int  j = ; j < n; j++)
        {
            if (strcmp(M_ID, stu[j].id) == ){//寻找这个同学的信息。
                print_max(stu[j].A_rank, stu[j].C_rank, stu[j].M_rank, stu[j].E_rank);
                flag = true;
                break;
            }
        }
        if (!flag) printf("N/A\n");

    }
    return ;
}

我这个可以改进一下，把id当int储存，题目中说了是6位数字。

并且后面处理查找ID的时候，复杂度比较高，算法笔记里面是申请了rank[10000000][4]这么大的数组。储存学生的排名信息，直接用ID去访问。但我总觉得申请这么大的内存，好虚呀，牺牲空间换时间。

其他我们的思路差不多。先储存，再分别排序，再对排序结果选择最高的输出。

这道题有个坑！

题目里面也没说，就是相同分数是怎么算排名。

比如91,90,88,88,84应该算作1,2,3,3,5。切记不要算作1,2,3,3,4。