1012 The Best Rank

Problem

Source: PAT 1012{target="_blank"}

Description

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

Sample Output

Solution

遍历过多会超时。。我剪枝了也超市，好像是N/A的判断所以用空间换时间，读取数据后进行四次遍历，计算出每人各自学科的排名，最后根据输入的Id输出四个学科里最优排名下的最有学科

Code

#include <iostream>
#include <algorithm>
#include <math.h>

using namespace std;
char out[4] = {'A', 'C', 'M', 'E'};
int index = 0;
int map[999999][4];
struct Student
{
    int id;
    int acme[4];
    void get_a()
    {
        acme[0] = (acme[3] + acme[1] + acme[2]) / 3;
    }
};
bool cmp(Student a, Student b)
{
    return a.acme[index] > b.acme[index];
}

main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int n, m, id;

    cin >> n >> m;
    Student stu[n];
    for (int i = 0; i < n; i++)
    {
        cin >> stu[i].id >> stu[i].acme[1] >> stu[i].acme[2] >> stu[i].acme[3];
        stu[i].get_a();
    }
    for (index = 0; index < 4; index++)
    {
        sort(stu, stu + n, cmp);
        map[stu[0].id][index] = 1;
        for (int j = 1; j < n; j++)
        {
            if (stu[j].acme[index] == stu[j - 1].acme[index])
            {
                map[stu[j].id][index] = map[stu[j - 1].id][index];
            }
            else
            {
                map[stu[j].id][index] = j + 1;
            }
        }
    }

    for (int i = 0; i < m; i++)
    {
        cin >> id;
        bool flag = false;
        for (int j = 0; j < n && !flag; j++)
        {
            if (stu[j].id == id)
            {
                flag = true;
            }
        }
        if (flag)
        {
            int min = *min_element(map[id], map[id] + 4);
            char res;
            for (int j = 0; j < 4; j++)
            {
                if (map[id][j] == min)
                {
                    res = out[j];
                    break;
                }
            }
            cout << min << " " << res << endl;
        }
        else
        {
            cout << "N/A" << endl;
        }
    }

    return 0;
}

1012 The Best Rank

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