PAT日志 1025

最新推荐文章于 2021-07-09 09:53:40 发布

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本文介绍了一个竞赛排名合并的问题，需要将多个地点的PAT竞赛排名列表合并，并生成最终的排名。通过比较分数和注册号，确保了同分选手排名一致，且按注册号排序。代码实现了排名的本地和全局排序。

顽强的小白

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

题目解析

排序题，比较简单，多场考试，要统计一个相同考场的排名，一个全部考生的排名，按照我不爱读题的习惯，看了输出结果直接就可以发现最后是按照全部考生的排名来输出的，所以先排各个考场的排名，排名储存在localRank中，最后排全部的，然后输出。

这里的排名方式仍然是PAT的老套路，是1 1 3 3 5 而不是1 1 2 2 3

代码实现

 if(a.score==b.score){
  return strcmp(a.id,b.id)<0;
 }else
  return a.score>b.score;
}
int main(){
 int n,k,u=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[u].id,&stu[u].score);
   stu[u++].local=i;
  }
  sort(stu+u-k,stu+u,cmp);
 
  int rank=1;
  int now=u-k;
  stu[now].localRank=rank;
  while(now<u-1){
   rank++;
   now++;
   if(stu[now].score==stu[now-1].score){
    stu[now].localRank=stu[now-1].localRank;
   }else{
    stu[now].localRank=rank; 
   }
  }
 }
 sort(stu,stu+u,cmp);
 int rank=1;
 int now=0;
 stu[now].finalRank=rank;
 while(now<u-1){
  rank++;
  now++;
  if(stu[now].score==stu[now-1].score){
   stu[now].finalRank=stu[now-1].finalRank;
  }else{
   stu[now].finalRank=rank; 
  }
 }
 printf("%d\n",u);
 for(int i=0;i<u;++i){
  printf("%s %d %d %d\n",stu[i].id,stu[i].finalRank,stu[i].local,stu[i].localRank);
 }
 return 0;
}