pat 1012

题目没说清楚等分数的情况怎么排名，我以为不用考虑--遂改，同等分数一个名次，否则名次就是分数比其高的人数+1。用了结构指针 数组，还有一个模板化的比较函数，模板参数是一个int，代表对应的ACME。开一个函数指针数组，放入四个具现化的comp。然后调用sort四次，每次把排名填入。ac- 

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <memory>
#include <map>

using namespace std;

struct stu {
    double scores[4];
    int rank[4];//A C M E
    string id;
    stu(const string& s, double c, double m, double e) :id(s)
    {
        scores[0] = (c + m + e) / 3;
        scores[1] = c, scores[2] = m, scores[3] = e;
    }
};

typedef bool(*comp)(shared_ptr<stu>&, shared_ptr<stu>&);
template<int i>
bool com(shared_ptr<stu>&lhs, shared_ptr<stu>& rhs)
{
    return lhs->scores[i] > rhs->scores[i];
}

pair<int, int> getpair(shared_ptr<stu>& ss)
{
    int best_rank = (1<<30);
    int best_sub = -1;
    for (int i = 0;i < 4;i++)
        if (ss->rank[i] < best_rank)
        {
            best_rank = ss->rank[i];
            best_sub = i;
        }
    return make_pair(best_sub, best_rank);
}
int main(void)
{
    vector<comp> com_vec;
    //for(int i=0;i<4;i++)
    com_vec.push_back(com<0>);
    com_vec.push_back(com<1>);
    com_vec.push_back(com<2>);
    com_vec.push_back(com<3>);
    

    const string subject = "ACME";
    int n, m;
    cin >> n >> m;
    vector<shared_ptr<stu> > sv;
    sv.reserve(n);
    for (int i = 0;i < n;i++)
    {
        string s;
        double c, m, e;
        cin >> s >> c >> m >> e;
        sv.push_back(shared_ptr<stu>(new stu(s, c, m, e)));
    }
    //sort 4 times
    for (int i = 0;i < 4;i++)
    {
        sort(sv.begin(), sv.end(), com_vec[i]);
        for (int j = 0;j < sv.size();j++)
        {
            //对应科目名次
            sv[j]->rank[i] = j + 1;
            if (j != 0 && sv[j]->scores[i] == sv[j - 1]->scores[i])
                sv[j]->rank[i] = sv[j - 1]->rank[i];
            
        }
    }
    typedef int sub;
    typedef int ran;
    map<string, pair<sub, ran> > ma;
    for (int i = 0;i < sv.size();i++)
    {
        auto p = getpair(sv[i]);
        //if (ma.find(sv[i]->id) == ma.end())
            ma[sv[i]->id] = p;
    }
    //give ans
    for (int i = 0;i < m;i++)
    {
        string temp;
        cin >> temp;
        auto it = ma.find(temp);
        if (it != ma.end())cout << it->second.second
            << " " << subject[it->second.first] << endl;
        else cout << "N/A" << endl;
    }

    return 0;
}