PAT甲级-排序类型-1025 PAT Ranking解题思路

1025 PAT Ranking (25 分)

思路

排序后，判断rank，可以借助循环，但是这题有个地方，就是内部需要进行多次排序，在读数据就排序可以大大简化代码量。

代码

#include <bits/stdc++.h>
using namespace std;

struct stu{
    
    
    char id[20];
    int score;
    int loc;
    int loc_rank;
}stud[300005];

bool cmp(stu a,stu b)
{
    
    
    if(a.score!=b.score)return a.score>b.score;
    else return strcmp(a.id,b.id)<0;
}
int main()
{
    
    
    int N;
    int num,all_num=0;
    cin>> N;

    int start = 0;

    for(int i =0;i<N;i++)
    {
    
    
        cin>>num;
        all_num+=num;
        for(int j=0;j<num;j++)
        {
    
    
            cin>>stud[start].id>>stud[start].score;
            stud[start].loc_rank = 0;
            stud[start++].loc = i+1;
        }
    }
    sort(stud,stud+start,cmp);
    cout<<all_num<<endl;

    int last_ranks[305]={
    
    1};
    int ranks[305]={
    
    0};
    int last_score[305]={
    
    0};
    for(int i =1;i<N+1;i++)
    {
    
    
        for(int j =0;j<start;j++)
        {
    
    
            if(stud[j].loc==i)
            {
    
    
                ranks[i]++;
                if(stud[j].score != last_score[i])
                {
    
    
                    last_score[i] =stud[j].score;
                    stud[j].loc_rank = ranks[i];
                    last_ranks[i] = ranks[i];
                }
                else 
                    stud[j].loc_rank = last_ranks[i];
            }
        }
    }

    
    int last_S = stud[0].score;
    int rank1 = 1;
    cout<<stud[0].id<< " "<< rank1 << " "<<stud[0].loc<<" "<<1<<endl;
    for(int i=1;i<start;i++)
    {
    
    
        if(stud[i].score == last_S)
        {
    
    
            cout<<stud[i].id<< " "<< rank1 << " "<<stud[i].loc<<" "<<stud[i].loc_rank<<endl;
        }
        else
        {
    
    
            cout<<stud[i].id<< " "<< i+1 << " "<<stud[i].loc<<" "<<stud[i].loc_rank<<endl;
            rank1 = i+1;
            last_S = stud[i].score;
        }
    }

}