PAT1012:The Best Rank

1012. The Best Rank (25)

時間限制

400 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

Standard

做者

CHEN, Yue

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

思路

  排序 + 查找。
  1.首先按順序輸入學生id和ACME四門課的成績。
  2.分別比較ACME四門成績的高低來決定學生的ACME的四個對應排名。
  3.找到每個學生中成績排名最高的那門課的索引，並將學生id和學生當前排名位置索引插入一個map中。
  4.輸出學生的最好成績的課程索引對應的最好排名和課程代號，若是學生id在map中不存在，輸出N/A。

代碼

#include<vector>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;

class student
{
public:
    student():myrank(4),grade(4)
    {

    }
    string id;
    int best_index;
    vector<int> myrank;
    vector<int> grade;
};
vector<char> grade_id = {'A','C','M','E'};
vector<student> students;
map<string,int> exists;
int index = 0;

bool cmp(const student& a,const student& b)
{
    return a.grade[index] > b.grade[index];
}


int main()
{
   int N,M;
   while(cin >> N >> M)
   {
      students.resize(N);
      for(int i = 0;i < N;i++)
      {
          cin >> students[i].id >> students[i].grade[1] >> students[i].grade[2] >> students[i].grade[3];
          students[i].grade[0] = (students[i].grade[1] + students[i].grade[2] + students[i].grade[3])/3;
      }

      //ACME sort
      for(;index <= 3;index++)
      {
          sort(students.begin(),students.end(),cmp);
          students[0].myrank[index] = 1;
          for(int i = 1;i < N;i++)
          {
              students[i].myrank[index] = i + 1;
              if(students[i - 1].grade[index] == students[i].grade[index])
              {
                 students[i].myrank[index] = students[i - 1].myrank[index];
              }
          }
      }

      //find best
      for(int i = 0;i < N;i++)
      {
          exists[students[i].id] = i;
          int minnum = students[i].myrank[0];
          for(int j = 1;j <= 3;j++)
          {
              if(students[i].myrank[j] < minnum)
              {
                  minnum = students[i].myrank[j];
                  students[i].best_index = j;
              }
          }
      }

      //query and output
      for(int i = 0;i < M;i++)
      {
          string id;
          cin >> id;
          if(exists.count(id) <= 0)
          {
              cout << "N/A" << endl;
          }
          else
          {
             int bindex = students[exists[id]].best_index;
             cout << students[exists[id]].myrank[bindex] << " " << grade_id[bindex] << endl;
          }
      }

   }
}