Pat class A - 1012 the best rank (PIT)

Link

The question ： Each student has his own serial number id, as well as C Language scene c, Math scores m, English scores e, And the average a. The title will output n Of a student id And three major subjects , Climate tracking m Time to ask , Ask someone id Students' best grades ranking and subject name .

Ideas ： The idea is very simple , Divide into four arrays to store the scores of all students in each subject, and then arrange them in order . But two pits need to be noticed ：1. The highest ranking of students must be selected according to a,c,m,e The priority of is the highest ranked account ( That is, each score is ranked third , But you must output 3 A).2. The same scores rank the same , But the next score will not be added up ( That is to say, now there is 90、90、89 Three achievements , The ranking of Qi is 1、1、3).

AC Code ：

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 1e5+10;

int n, m;
struct node{
    int c, m, e, a;
};

map<string, node> mp;
vector<node> cc, mm, ee, aa;
map<int, int> gc, gm, ge, ga;
map<string, int> vis;

bool cmpc(node x, node y){
    return x.c>y.c;
}
bool cmpm(node x, node y){
    return x.m>y.m;
}
bool cmpe(node x, node y){
    return x.e>y.e;
}
bool cmpa(node x, node y){
    return x.a>y.a;
}

signed main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i ++){
        string s; int x, y, z;
        cin >> s >> x >> y >> z;
        vis[s] = 1;
        node tmp = {x, y, z, (x+y+z)/3};
        mp[s] = tmp;
        cc.push_back(tmp);
        mm.push_back(tmp);
        ee.push_back(tmp);
        aa.push_back(tmp);
    }

    sort(cc.begin(), cc.end(), cmpc);
    sort(mm.begin(), mm.end(), cmpm);
    sort(ee.begin(), ee.end(), cmpe);
    sort(aa.begin(), aa.end(), cmpa);

    for(int i = 0; i < cc.size(); i ++){
        if(!gc[cc[i].c]) gc[cc[i].c] = i+1;
    }
    for(int i = 0; i < mm.size(); i ++){
        if(!gm[mm[i].m]) gm[mm[i].m] = i+1;
    }
    for(int i = 0; i < ee.size(); i ++){
        if(!ge[ee[i].e]) ge[ee[i].e] = i+1;
    }
    for(int i = 0; i < aa.size(); i ++){
        if(!ga[aa[i].a]) ga[aa[i].a] = i+1;
    }

    while(m--){
        string s;
        cin >> s;
        if(!vis[s]){
            cout << "N/A" << endl;
            continue;
        }
        int pc = gc[mp[s].c], pm = gm[mp[s].m], pe = ge[mp[s].e], pa = ga[mp[s].a];
        if(pa<=pc&&pa<=pm&&pa<=pe) cout << pa << " A" << endl;
        else if(pc<=pm&&pc<=pe&&pc<=pa) cout << pc << " C" << endl;
        else if(pm<=pc&&pm<=pe&&pm<=pa) cout << pm << " M" << endl;
        else if(pe<=pc&&pe<=pm&&pe<=pa) cout << pe << " E" << endl;
//        cout << mp[s].c << " " << mp[s].m << " " << mp[s].e << " " << mp[s].a << endl;
//        cout << pc << " " << pm << " " << pe << " " << pa << endl << endl;
    }

    return 0;
}