PAT Advanced 1025. PAT Ranking (25) (C语言实现)

我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到Github Pages浏览最新内容(本篇文章链接)。此处文章目前已更新至与Github Pages同步。欢迎star我的repo。

题目

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 $N$ ( $\le 100$ ), the number of test locations. Then $N$
ranklists follow, each starts with a line containing a positive integer $K$ (
$\le 300$ ), 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

一道有关于排序的题目。题目要求得到每个考生分排名和总排名，因此关键点就在于如何
排序。

思路：由于读取是以考点为组，因此可以在每读完一个考点，即对此考点内的考生进
行排序，得到分排名。最后再对整个列表进行排序，得到总排名。

实现：

使用一个数组存储。排序时，使用qsort传入不同位置和长度参数即可。

代码

最新代码@github，欢迎交流

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

/* registration number, location nubmer, total score, final rank, local rank */
typedef struct {
    char reg[14];
    int loc, score, rank, lrank;
} Testee, *pTestee;

int cmp(const void *a, const void *b)
{
    pTestee p1 = *(pTestee*)a, p2 = *(pTestee*)b;
    if(p1->score != p2->score)
        return p2->score - p1->score;
    else
        return strcmp(p1->reg, p2->reg);
}

int main()
{
    int N, K;
    Testee buffers[100 * 300] = {0}, *q = buffers;
    pTestee testees[100 * 300] = {0}, *p = testees;

    scanf("%d", &N);
    for(int n = 0; n < N; n++)
    {
        /* read one test location */
        scanf("%d", &K);
        for(int k = 0; k < K; k++, p++)
        {
            *p = q++;  /* use a new cell */
            (*p)->loc = n + 1;
            scanf("%s %d", (*p)->reg, &(*p)->score);
        }

        /* sort the kth location sublist */
        qsort(p - K, K, sizeof(pTestee), cmp);

        /* get the local rank, i is the offset (we need left K cells of p) */
        int lrank = 1;
        pTestee *lp;
        (*(p - K))->lrank = lrank;  /* boundary special case */
        for(int i = 1; i < K; i++)
        {
            lp = p - K + i;
            if((*lp)->score != (*(lp - 1))->score)
                lrank = i + 1;
            (*lp)->lrank = lrank;
        }
    }

    /* sort the whole list */
    qsort(testees, p - testees, sizeof(pTestee), cmp);
    /* get the final rank */
    int rank = 1;
    pTestee *fp;
    (*testees)->rank = rank;
    for(int i = 1; testees + i < p; i++)
    {
        fp = testees + i;
        if((*fp)->score != (*(fp - 1))->score)
            rank = i + 1;
        (*fp)->rank = rank;
    }

    /* output */
    printf("%ld\n", p - testees);
    for(pTestee* i = testees; i < p; i++)
        printf("%s %d %d %d\n", (*i)->reg, (*i)->rank, (*i)->loc, (*i)->lrank);

    return 0;
}