1012 The Best Rank （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 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:

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：根据题意，可设置rank[][] 数组来存储每个编号在按照（A，C，M，S）排序规则下的4中不同排名，然后输出每个排名中最小的排名， 注意在排名的时候，若和上一个分数相同，则排名保持不变，但是会占用掉一个名额，下个分数低的 将以已经排名人数为基础，向下递增。（其实在程序中表达出来就是其下标i+1）

 

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
#define ll long long
#define vi vector<int>
#define pii pair<int, int>
#define x first
#define y second
const int maxn=2100;
const int maxr=1000000;
struct stu
{
    int id;
    int grade[4];
}st[maxn];
char course[4]={'A','C','M','E'};
int n,m,now,rankstu[1000000][4] = {0};;
bool cmp(stu s1, stu s2)
{
    return s1.grade[now]>s2.grade[now];
}
int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
        cin>>st[i].id>>st[i].grade[1]>>st[i].grade[2]>>st[i].grade[3];
        st[i].grade[0]=round((st[i].grade[1]+st[i].grade[2]+st[i].grade[3])/3.0)+0.5;
    }
    for(now=0;now<4;now++)
    {
        sort(st,st+n,cmp);
        rankstu[st[0].id][now] = 1;
        for(int i=1;i<n;i++)
        {
            if(st[i].grade[now]==st[i-1].grade[now])
            {
                 rankstu[st[i].id][now]=rankstu[st[i-1].id][now];
            }
            else{

                 rankstu[st[i].id][now]=i+1;
            }
        }
    }
    /*
    for(int i=0;i<4;i++)
    {
        cout<<rankstu[310101][i]<<endl;
    }
    */
    int qu;
    for(int i=0;i<m;i++)
    {
        cin>>qu;
        if(rankstu[qu][0]==0)
        {
            cout<<"N/A"<<endl;
        }
        else
        {
            int cnt=0,inf=maxn;
            for(int j=0;j<4;j++)
            {
                if(rankstu[qu][j]<inf)
                {
                    inf=rankstu[qu][j];
                    cnt=j;
                }
            }
            cout<<rankstu[qu][cnt]<<" "<<course[cnt]<<endl;

        }
    }
    return 0;
}