题目描述

简单的排序模拟，也没啥细节，想明白了写就好了。
每个区域处理一遍排序和排名，然后再全体处理一遍排序和排名就行了。

时间复杂度

复杂度瓶颈在排序上，是$O(nlogn)$。

C++ 代码

#include <iostream>
#include <algorithm>
using namespace std;
int n, k, c;
struct S
{
    string registration_number;
    int final_rank;
    int location_number;
    int local_rank;
    int score;
    bool operator<(S s)
    {
        if (score != s.score) return score >= s.score;
        return registration_number < s.registration_number;
    }
}s[30010];
void gen_local_rank(S* l, S* r)
{
    int num = 2;
    int rank = 1;
    S* i = l + 1;
    l -> local_rank = 1;
    while (i != r)
    {
        if ((i - 1) -> score == i -> score)
            i -> local_rank = rank;
        else
            i -> local_rank = rank = num;
        num ++, i ++;
    }
}
void gen_final_rank(S* l, S* r)
{
    int num = 2;
    int rank = 1;
    S* i = l + 1;
    l -> final_rank = 1;
    while (i != r)
    {
        if ((i - 1) -> score == i -> score)
            i -> final_rank = rank;
        else
            i -> final_rank = rank = num;
        num ++, i ++;
    }
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
    {
        cin >> k;
        for (int j = 0; j < k; j ++)
        {
            string rn;
            int sc;
            cin >> rn >> sc;
            s[c ++] = {rn, 0, i, 0, sc};
        }
        sort(s + c - k, s + c);
        gen_local_rank(s + c - k, s + c);
    }
    sort(s, s + c);
    gen_final_rank(s, s + c);
    cout << c << endl;
    for (int i = 0; i < c; i ++)
        cout << s[i].registration_number << " " << s[i].final_rank << " " << s[i].location_number << " "<< s[i].local_rank << endl;
}