PAT 1025 PAT Ranking（25 分）（排序+排名）

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最新推荐文章于 2025-06-24 21:07:38 发布

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博客围绕PAT排名合并问题展开，介绍了编程能力测试PAT由浙江大学计算机学院组织，需编写程序合并各考点排名生成最终排名。给出了输入、输出规格及示例，题解提醒处理分数相同排名相同情况，并给出代码。

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

题解：注意处理分数相同排名相同的情况。

代码：

#include<bits/stdc++.h>
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()
{
    int n,k,num=0;
    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++;
        }
        sort(stu+num-k,stu+num,cmp);  //重点
        stu[num-k].local_rank=1;      //分数最高的排名第一
        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;
            else
                stu[j].local_rank=j+1-(num-k);
        }
    }
    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;
}