PAT The Best Rank[未作]

1012 The Best Rank (25)（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 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

//emm,这个没写出来，看了别人的之后恍然大悟，我应该过几天之后再自己写出来。

 代码来自:https://www.liuchuo.net/archives/2207

#include <cstdio>
#include <algorithm>
using namespace std;
struct node {
    int id, best;
    int score[4], rank[4];
}stu[2005];
int exist[1000000], flag = -1;//厉害了，不用map，用哈希数组。
bool cmp1(node a, node b) {return a.score[flag] > b.score[flag];}
//我的天，这个flag可以默认传进来。。
int main() {
    int n, m, id;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < n; i++) {
        scanf("%d %d %d %d", &stu[i].id, &stu[i].score[1], &stu[i].score[2], &stu[i].score[3]);
        stu[i].score[0] = (stu[i].score[1] + stu[i].score[2] + stu[i].score[3]) / 3.0 + 0.5;
        //要四舍五入，所以+0.5后取整。
    }
    for(flag = 0; flag <= 3; flag++) {
        sort(stu, stu + n, cmp1);
        //有可能按照这个排出来之后，连续好几个数都相等。
        stu[0].rank[flag] = 1;
        for(int i = 1; i < n; i++) {
            stu[i].rank[flag] = i + 1;//理论上来说先赋值为i+1,再判断和前边相等，那么就再更新。
            if(stu[i].score[flag] == stu[i-1].score[flag])
                stu[i].rank[flag] = stu[i-1].rank[flag];//那么就直接赋值相等就可以了。
        }
    }
    for(int i = 0; i < n; i++) {
        exist[stu[i].id] = i + 1;
        stu[i].best = 0;
        int minn = stu[i].rank[0];
        for(int j = 1; j <= 3; j++) {//这个也是通过循环找到。
            if(stu[i].rank[j] < minn) {
                minn = stu[i].rank[j];
                stu[i].best = j;
            }
        }
    }
    char c[5] = {'A', 'C', 'M', 'E'};
    for(int i = 0; i < m; i++) {
        scanf("%d", &id);
        int temp = exist[id];
        if(temp) {
            int best = stu[temp-1].best;
            printf("%d %c\n", stu[temp-1].rank[best], c[best]);
        } else {
            printf("N/A\n");
        }
    }
    return 0;
}

//看这个代码受到了很多启发：1.如果出现分数名相等的情况，该怎么处理？for循环写的非常好，并且根据题目特点使用了哈希数组存储id对应的下标，但是因为牛客网上的id中包含字母，所以就通不过，这个地方可以改进，明天我来试一试。