PAT(A) 1025. PAT Ranking (25)

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:

2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85

Sample Output:

9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=30005;

struct Node{
    char id[13];            //准考证号
    int score;              //分数
    int location_number;    //考场号
    int location_rank;      //考场内排名
}stu[maxn];

bool cmp(Node a, Node b){
    if(a.score!=b.score)                //先按分数从高到低排序
        return a.score>b.score;
    else return strcmp(a.id, b.id)<0;   //分数相同按准考证号从小到大排序
}

int main()
{
    int n, k, cnt=0;    //n为考场数，k为当前考场内的人数，cnt为总考生数
    scanf("%d", &n);
    for(int i=0; i<n; i++){
        scanf("%d", &k);
        cnt += k;
        for(int j=0; j<k; j++){
            stu[cnt-k+j].location_number=i+1;   //该考生的考场号为 i+1
            scanf("%s%d", &stu[cnt-k+j].id, &stu[cnt-k+j].score);
        }
        sort(stu+cnt-k, stu+cnt, cmp);          //将该考场的考生排序
        //location_rank
        stu[cnt-k].location_rank=1;             //该考场第1名的 location_rank 记为1
        for(int j=cnt-k+1; j<cnt; j++){         //对该考场剩余的考生存 location_rank
            //如果与前一位考生同分，则 location_rank 也相同
            if(stu[j].score==stu[j-1].score)        
                stu[j].location_rank=stu[j-1].location_rank;
            //如果与前一位考生不同，则 location_rank 为该考生前的人数
            else
                stu[j].location_rank=j+1-(cnt-k);  
         }
    }
    printf("%d\n", cnt);        //输出总考生数
    sort(stu, stu+cnt, cmp);    //将所有考生排序
    //all_rank
    int all_rank=1;             //设当前的总排名allrank
    for(int i=0; i<cnt; i++){
        //当前考生与上一个考生分数不同时，让 allrank=i+1（前面的人数+1）
        if(i>0 && stu[i].score!=stu[i-1].score) 
            all_rank=i+1;
        printf("%s ", stu[i].id);
        printf("%d %d %d\n", all_rank, stu[i].location_number, stu[i].location_rank);
    }
    return 0;
}
    

 

转载于:https://www.cnblogs.com/claremore/p/6549526.html