【PAT A1025】PAT Ranking

PAT Ranking

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

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805474338127872

题解：

1.首先读入考生信息，按照考场来读入；读入一个考场考生信息后就对该考场考生进行排序，并将排名结果存入其结构体中。
2.对所有考生进行排序。
3.按顺序一边计算总排名，一边输出所有考生的信息。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct Student{
    char id[15];
    int score;
    int location_number;
    int local_rank;
}stu[300010];

bool cmp(Student a,Student b){
    if(a.score != b.score)  return a.score > b.score;   //分数从高到低排序
    else return strcmp(a.id,b.id)<0;    //再按照学生准考证号
}

int main(){
    int n,k,num = 0;    //num为总考生人数
    scanf("%d", &n);    //n表示考场数目
    for(int i=1;i<=n;i++){
        scanf("%d", &k);    //当前考场人数
        for(int j=0;j<k;j++){
            scanf("%s %d",stu[num].id, &stu[num].score);
            stu[num].location_number = i;
            num++;
        }
        sort(stu+num-k, stu+num, cmp);
        stu[num - k].local_rank = 1;
        for(int j=num-k+1; j<num; j++){
            if(stu[j].score != stu[j-1].score)
//                 stu[j].local_rank = stu[j-1].local_rank + 1;
//                 上面的写法是错误的！加入出现两个人分数一样，那排名就一样，但是你后面的学生就不一定是前面学生排名加1了，而应该加2才对！
                stu[j].local_rank = j-(num-k)+1;
            else
                stu[j].local_rank = stu[j-1].local_rank;
        }
    }
    printf("%d\n",num);
    sort(stu, stu+num, cmp);
    int rank = 1;
    for(int i=0;i<num;i++){
        if(i>0 && stu[i].score != stu[i-1].score)
            rank = i+1;
        printf("%s ",stu[i].id);
        printf("%d %d %d\n",rank,stu[i].location_number,stu[i].local_rank);
    }
    return 0;
}

__EOF__

本文作者：techgy
本文链接：https://www.cnblogs.com/techgy/p/15032774.html
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