1025 PAT Ranking (25分)

1025 PAT Ranking (25分)

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

思路很简单，先分考场排序然后再一起排序

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct Student{
    char id[15];
    int score;             
    int location_number;     //考场号
    int local_rank;         //考场内排名
}stu[30010];

bool cmp(Student a,Student b)
{
    if(a.score != b.score)
        return a.score > b.score;
    else
        return strcmp(a.id,b.id) < 0;
}

int main(void)
{
    int n,k,num = 0;          //n为考场数，k为考场人数，num来计算学生总数
    scanf("%d", &n);
    for(int i = 1;i <= n; i++){
        scanf("%d", &k);
        for(int j = 0; j < k; j++){
            scanf("%s %d", stu[num].id,&stu[num].score);
            stu[num].location_number = i;
            num++;
        }            //读取数据
        sort(stu+num-k,stu+num,cmp);    //考场内排序，对单个考场k个人单独排序
        stu[num-k].local_rank = 1;
        for(int j = num - k + 1; j < num; j++){  //从第二名开始循环，不然j-1会干扰到前一个考场
            if(stu[j].score == stu[j-1].score){
                stu[j].local_rank = stu[j-1].local_rank;  //分数一样，排名一样
            }
            else{
                stu[j].local_rank = j+1-(num-k);     //不是前一名+1！！！！
            }
        }
    }
    printf("%d\n",num);
    sort(stu,stu+num,cmp);
    int r = 1;
    for(int i = 0; i < num; i++){
        if(i > 0 && stu[i].score != stu[i-1].score){
            r = i+1;
        }
        printf("%s ",stu[i].id);
        printf("%d %d %d\n",r,stu[i].location_number,stu[i].local_rank);
    }
    return 0;
}

来源：CSDN

作者：Luyaooooo0

链接：https://blog.csdn.net/Luyaooooo0/article/details/104803057