1012. The Best Rank (25)

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

题目要点：这里只需注意出现并列名次的时候，并列名次的排名一样，但是后一个的排名是按照其原来对应的名次。举个栗子：要表示1、1、2、3、4排名，则这个排名应该是1、1、3、4、5。

#include <stdio.h>
#include <stdlib.h>

struct people
{
    int id, best;
    int score[4], rank[4];

}p[2001];

int f;

int cmp(const void *a,const void *b)
{
     struct people *aa=(struct people *)a;
     struct people *bb=(struct people *)b;
     return bb->score[f] - aa->score[f];
}

int main()
{
    int visited[1000000] = {0}; //在codeblock上只能创建100000的数组。
    char project[4] = {'A', 'C', 'M', 'E'}; //每个科目的代号
    int n, m, i, id, min, tmp;
    scanf("%d %d", &n, &m);
    for(i = 0; i < n; i++) //输入和存储数据，注意分数的存放顺序
    {
        scanf("%d %d %d %d", &p[i].id, &p[i].score[1], &p[i].score[2], &p[i].score[3]);
        p[i].score[0] = (p[i].score[1] + p[i].score[2] + p[i].score[3]) / 3;
    }
    for(f = 0; f <= 3; f++)//实现纵向排序，选出每一个人的每个科目的排名
    {
        qsort(p, n, sizeof(p[0]), cmp);
        p[0].rank[f] = 1;
        for(i = 1; i < n; i++)
        {
            p[i].rank[f] = i + 1;
            if(p[i].score[f] == p[i - 1].score[f])
                p[i].rank[f] = p[i - 1].rank[f];
        }
    }
    for(i = 0; i < n; i++) //选出排名最好的科目的下标
    {
        visited[p[i].id] = i + 1; //记录每个人的下标，类似散列表
        //+1是为了与不存在的元素用0表示相区别
        p[i].best = 0;
        min = p[i].rank[0];
        for(f = 1; f <= 3; f++)
        {
            if(p[i].rank[f] < min)
            {
                min = p[i].rank[f];
                p[i].best = f;
            }
        }
    }
    for(i = 0; i < m; i++)//实现输出
    {
        scanf("%d", &id);
        tmp = visited[id];
        if(tmp)
        {
            int t = p[tmp - 1].best;
            printf("%d %c\n", p[tmp - 1].rank[t], project[t]);
        }
        else
            printf("N/A\n");
    }

    return 0;
}