PAT甲级——PAT Judge

题目描述

The ranklist of PAT is generated from the status list, which shows the scores of the submittions.  This time you are supposed to generate the ranklist for PAT.

 

输入描述:

Each input file contains one test case.  For each case, the first line contains 3 positive integers, N (<=104), the total number of users, K (<=5), the total number of problems, and M (<=105), the total number of submittions.  It is then assumed that the user id's are 5-digit numbers from 00001 to N, and the problem id's are from 1 to K.  The next line contains K positive integers p[i] (i=1, ..., K), where p[i] corresponds to the full mark of the i-th problem.  Then M lines follow, each gives the information of a submittion in the following format:
user_id problem_id partial_score_obtained
where partial_score_obtained is either -1 if the submittion cannot even pass the compiler, or is an integer in the range [0, p[problem_id]].  All the numbers in a line are separated by a space.

 

输出描述:

For each test case, you are supposed to output the ranklist in the following format:
rank user_id total_score s[1] ... s[K]
where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem.  If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position.  If a user has submitted several solutions to solve one problem, then the highest score will be counted.
The ranklist must be printed in non-decreasing order of the ranks.  For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems.  And if there is still a tie, then they must be printed in increasing order of their id's.  For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist.  It is guaranteed that at least one user can be shown on the ranklist.

 

输入例子:

• 7 4 20
  20 25 25 30
  00002 2 12
  00007 4 17
  00005 1 19
  00007 2 25
  00005 1 20
  00002 2 2
  00005 1 15
  00001 1 18
  00004 3 25
  00002 2 25
  00005 3 22
  00006 4 -1
  00001 2 18
  00002 1 20
  00004 1 15
  00002 4 18
  00001 3 4
  00001 4 2
  00005 2 -1
  00004 2 0

 

输出例子:

1 00002 63 20 25 - 18
2 00005 42 20 0 22 -
2 00007 42 - 25 - 17
2 00001 42 18 18 4 2
5 00004 40 15 0 25 -

 

注意点：

• 首先把考生各科目的成绩初始化为-1，当输入成绩不为-1时，将成绩进行替换。如果后面有同一科目更大的成绩输入进来，则要再次进行提换。如果输入的成绩第一次等于满分则要使完美解题数加一，这里要注意设置判断是否为第一次达到满分成绩。

  #include 
  #include
  #include
  using namespace std;
  const int maxn = 10010;
  
  struct Student {
  	int id;
  	int score[6];
  	bool flag;
  	int score_all;
  	int solve;
  }stu[maxn];
  
  int  n, k, m;
  int full[6];
  bool cmp(Student a, Student b)
  {
  	if (a.score_all != b.score_all) return a.score_all > b.score_all;
  	else if (a.solve != b.solve) return a.solve > b.solve;
  	else return a.id < b.id;
  }
  
  void init() {
  	for (int i = 1; i <= n; i++)
  	{
  		stu[i].id = i;
  		stu[i].score_all = 0;
  		stu[i].solve = 0;
  		stu[i].flag = false;
  		memset(stu[i].score, -1, sizeof(stu[i].score));
  	}
  }
  int main()
  {
  	scanf("%d%d%d", &n, &k, &m);
  	init();
  	for (int i = 1; i <= k; i++)
  	{
  		scanf("%d", &full[i]);
  	}
  	int u_id, p_id, score_obtained;
  	for (int i = 0; i < m; i++)
  	{
  		scanf("%d%d%d", &u_id, &p_id, &score_obtained);
  		//stu[u_id].id = u_id;
  		if (score_obtained != -1) {
  			stu[u_id].flag = true;
  		}
  		if (score_obtained == -1 && stu[u_id].score[p_id] == -1)
  		{
  			stu[u_id].score[p_id] = 0;
  		}
  		if (score_obtained == full[p_id] && stu[u_id].score[p_id]  stu[u_id].score[p_id]) {
  			stu[u_id].score[p_id] = score_obtained;
  		}
  
  	}
  	for (int i = 1; i <= n; i++)
  	{
  		for (int j = 1; j <= k; j++)
  		{
  			if (stu[i].score[j] != -1)
  			{
  				stu[i].score_all += stu[i].score[j];
  			}
  				
  		}
  	}
  	sort(stu + 1, stu + n + 1, cmp);
  	int r=1;
  	//printf_s("%d %d %d\n", stu[2].score[2], stu[2].score[3], stu[2].score[4]);
  	for (int i = 1; i <= n && stu[i].flag==true; i++)
  	{
  		if (i>1&&stu[i].score_all != stu[i - 1].score_all) r = i;
  		printf("%d %05d %d", r, stu[i].id, stu[i].score_all);
  		for (int j = 1; j <= k; j++)
  		{
  			if (stu[i].score[j] != -1) printf(" %d", stu[i].score[j]);
  			else printf(" -");
  		}
  		printf("\n");
  		
  			
  	}
  	return 0;
  }