【PAT甲级】1012 The Best Rank

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📝原题地址：题目详情 - 1012 The Best Rank (pintia.cn)
🔑中文翻译：最佳排名
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1012 The Best Rank

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
1
2
3
4
5

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
1
2
3
4
5
6
7
8
9
10
11
12

Sample Output:

1 C
1 M
1 E
1 A
3 A
N/A
1
2
3
4
5
6

题意

我们只考虑他们的三门成绩：C–c语言，M–数学，E–英语，除此之外，我们还会考虑 A –三门成绩平均值。

注意： 平均成绩为三科成绩平均值四舍五入取整的结果。

我们要输出每个学生四门成绩中排名最高的那一科，如果遇到排名相同的，则按照 $A>C>M>E$ 规则输出他的最佳排名。

如果无法查询到该学生的成绩，则输出 N/A。

思路

1. 首先，我们用到了两个容器对学生的信息进行存储。一个是哈希表 grades ，能够通过学生的 id 快速查找到其对应的成绩。另一个是二维数组 q ，从 0~4 层数组分别存储 ACME 的成绩。

2. 对 q 中的四层数组进行排序，这样就得到了每一科目包括平均值从小到大的排序。

3. 开始查找对应学生的最佳科目排名，我们这里采用的是二分查找法，因为 q 中已经是有序的了，所以可以二分找到对应的排名。

   这里需要注意的是由于数组中是从小到大进行排序的，所以下标越大分数越高，但是我们输出的顺序是反过来的，排名序号越小代表分数越高。所以这里的二分查找需要找到分数的右边界，然后再用 g.szie() 减去它就是最终的排名。

   例如，86,88,88,90,91 这五个成绩，我们要查找的是 88 ，而 88 有两个，二分查找需要找到 88 的右边界即上面例子中下标为 2 的数字（下标从 0 开始），然后再用 g.szie() 减去它得到 5 - 2 = 3 ，也就是排名第 3 ，正好符合想要的结果。

4. 最后要输出结果，字符 ACME 可以用一个数组存起来，方便调用。另外，如果在哈希表找不到给定 id 需要输出 N/A 。

代码

#include
using namespace std;

unordered_map<string, vector<int>> grades;
vector<int> q[4];   //q[0]:A,q[1]:C,q[2]:M,q[3]:E

//二分获取排名
int get_rank(vector<int> g, int x)
{
    int l = 0, r = g.size() - 1;
    while (l < r)
    {
        int mid = (l + r + 1) >> 1;
        if (g[mid] <= x)    l = mid;
        else    r = mid - 1;
    }
    return g.size() - r;
}

int main()
{
    int n, m;
    cin >> n >> m;

    //输入学生信息
    for (int i = 0; i < n; i++)
    {
        string id;
        int t[4] = { 0 };
        cin >> id;
        for (int j = 1; j < 4; j++)
        {
            cin >> t[j];
            t[0] += t[j];
        }
        t[0] = round(t[0] / 3.0);   //求平均值（四舍五入）
        for (int j = 0; j < 4; j++)
        {
            grades[id].push_back(t[j]);
            q[j].push_back(t[j]);
        }
    }

    //对成绩进行排序
    for (int i = 0; i < 4; i++)    sort(q[i].begin(), q[i].end());

    //查找学生成绩排名
    char names[4] = { 'A','C','M','E' };
    while (m--)
    {
        string id;
        cin >> id;
        //判断id是否存在
        if (!grades.count(id))   puts("N/A");
        else
        {
            char c;
            int minv = 1e9;
            for (int i = 0; i < 4; i++)
            {
                int rank = get_rank(q[i], grades[id][i]);
                //排名相等情况下，A>C>M>E
                if (rank < minv)
                {
                    minv = rank;
                    c = names[i];
                }
            }
            cout << minv << " " << c << endl;
        }
    }

    return 0;
}
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