1012 The Best Rank （25 point(s)）

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

 大意：

给出　Ｎ个数据　和Ｍ　个排名请求

每个请求，输入对应的ｉｄ号，返回AMEC 中最好的排名，当然排名有优先级

１．使用了四个vector　去放置对应的ｉｄ和对应的成绩

２.然后对其进行排序操作，从大到小

３．然后在用unordered_map 进行每一个id　的存储，存储的是当前的最好成绩

注意点：

unordered_map　中的struct 中必须包含默认的参数初始化构造方法

一定要注意成绩并列的情况，比如两个Ｃ中都排名第一，那么就是排名两个的名次都是第一

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

struct Anode{
    string name;
    float grade;
    Anode(string str,float grade):name(str),grade(grade){}
};

struct Cnode{
    string name;
    int grade;
    Cnode(string str,int grade):name(str),grade(grade){}
};
struct Mnode{
    string name;
    int grade;
    Mnode(string str,int grade):name(str),grade(grade){}
};
struct Enode{
    string name;
    int grade;
    Enode(string str,int grade):name(str),grade(grade){}
};

struct Outer{
    char tag;
    int level;
    Outer(char tag = ' ', int level =0):tag(tag),level(level){}
};


bool Acmp(const Anode &a,const Anode &b){
    return a.grade>b.grade;
}

bool Ccmp(const Cnode &a,const Cnode &b){
    return a.grade>b.grade;
}

bool Mcmp(const Mnode &a,const Mnode &b){
    return a.grade>b.grade;
}

bool Ecmp(const Enode &a,const Enode &b){
    return a.grade>b.grade;
}



int main(){

    int N,M;
    cin>>N>>M;

    vector<Anode> Alist;
    vector<Cnode> Clist;
    vector<Mnode> Mlist;
    vector<Enode> Elist;
    unordered_map<string,Outer> out;

    for(int i=0;i<N;i++){
        string name_temp;
        int grade =0;
        cin>>name_temp>>grade;
        Clist.push_back(Cnode(name_temp,grade));
        cin>>grade;
        Mlist.push_back(Mnode(name_temp,grade));
        cin>>grade;
        Elist.push_back(Enode(name_temp,grade));
        Alist.push_back(Anode(name_temp,((float)(grade+Clist[i].grade+Mlist[i].grade)/3.0)));
    }

    sort(Alist.begin(),Alist.end(),Acmp);
    sort(Clist.begin(),Clist.end(),Ccmp);
    sort(Mlist.begin(),Mlist.end(),Mcmp);
    sort(Elist.begin(),Elist.end(),Ecmp);


    for(int i=0;i<N;i++){

        int level_ = i+1;
        if(i>0&&Elist[i].grade==Elist[i-1].grade){
            level_=i;
        }

        if(out.find(Elist[i].name)!=out.end()){

            if(out[Elist[i].name].level>=level_){
                Outer  temp('E',level_);
                out[Elist[i].name]= temp;
            }
        }else{
            Outer  temp('E',level_);
            out[Elist[i].name]= temp;

        }
    }



    for(int i=0;i<N;i++){

        int level_ = i+1;
        if(i>0&&Mlist[i].grade==Mlist[i-1].grade){
            level_=i;
        }

        if(out.find(Mlist[i].name)!=out.end()){
            if(out[Mlist[i].name].level>=level_){

                Outer  temp('M',level_);
                out[Mlist[i].name]= temp;
            }
        }else{

            Outer  temp('M',level_);
            out[Mlist[i].name]= temp;
        }
    }


    for(int i=0;i<N;i++){
        int level_ = i+1;
        if(i>0&&Clist[i].grade==Clist[i-1].grade){
            level_=i;
        }
        if(out.find(Clist[i].name)!=out.end()){
            if(out[Clist[i].name].level>=level_){

                Outer  temp('C',level_);
                out[Clist[i].name]= temp;
            }
        }else{

            Outer  temp('C',level_);
            out[Clist[i].name]= temp;
        }
    }

    for(int i=0;i<N;i++){
        int level_ = i+1;
        if(i>0&&Alist[i].grade==Alist[i-1].grade){
            level_=i;
        }
        if(out.find(Alist[i].name)!=out.end()){
            if(out[Alist[i].name].level>=level_){

                Outer  temp('A',level_);
                out[Alist[i].name]= temp;
            }
        }else{

            Outer  temp('A',level_);
            out[Alist[i].name]= temp;
        }
    }


    for(int i=0;i<M;i++){
        string name_;
        cin>>name_;
        if(out.find(name_)!=out.end()){
            cout<<out[name_].level<<" "<<out[name_].tag<<endl;
        }else{
            cout<<"N/A"<<endl;
        }


    }

    return 0;
}

calculus

n. [病理] 结石；微积分学

linear algebra 线性代数

emphasizing

v. 强调（emphasize的现在分词）