PAT甲级1075 PAT Judge//多练习

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

Input Specification:

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 submissions. 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 submission in the following format:

user_id problem_id partial_score_obtained

where partial_score_obtained is either −1 if the submission 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.

Output Specification:

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.

Sample Input:

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

Sample Output:

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 -

思路

这题核心考察排序，大致的要求是按分数从高到低排序，但是细小的要求很多。
一个是要记录学生的题目通过率，也就是有几个题目拿到了满分
注意这句话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.也就是不能输出那些一个都没编译通过的或从来没提交过的人。所以这里将score[i]初始化为-1，表示该题没有提交过；当题出现无法通过编译的提交时得分记为0，表示题目有过提交

• 题目还有一个很隐晦的边界情况：某个考生所有题目都通过了编译，但是都得了0分。这个学生也是要输出的，因为并没有违背“全场没有提交，或是没有能通过编译的提交”这个情况。他编译通过了，也提交了，只是都得了0分而已。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct usr
{
	int id;
	int score[6];
	int solved;
	int total;
	bool flag;
	int rank;
	usr()
	{
		flag = false;
		memset(score, -1, sizeof(score));
		solved = 0;
		total = 0;
		rank = 0;
	}
}student[10010];
int n, k, m;
int full[6];
bool cmp(usr a, usr b)
{
	if (a.total != b.total)
		return a.total > b.total;
	else if (a.solved != b.solved)
		return a.solved > b.solved;
	else
		return a.id < b.id;
}
void output(usr a)
{
	printf("%d %05d %d", a.rank, a.id, a.total);
	for (int i = 1; i <= k; i++)
	{
		if (a.score[i] == -1)
			cout << " -";
		else
			cout << " " << a.score[i];
	}
	cout << endl;
}
int main()
{
	cin >> n >> k >> m;
	for (int i = 1; i <= k; i++)
		cin >> full[i];
	for (int i = 0; i < m; i++)
	{
		int stu, prob, mark;
		cin >> stu >> prob >> mark;
		student[stu].id = stu;
		if (mark != -1) //flag用来标记是否有通过编译的题目
			student[stu].flag = true; //最后不会输出flag = false的学生
		if (mark == -1 && student[stu].score[prob] == -1)
		//坑点
		//如果是第一次编译，分值记为0分
			student[stu].score[prob] = 0;
		if (mark == full[prob] && student[stu].score[prob] < mark)
		//坑点
		//如果不带&& student[stu].score[prob] < mark的判断，学生第二次满分提交solved会再次增加，但只要记第一次的
		{
			student[stu].score[prob] = mark;
			student[stu].solved++;
		}
		else if (mark > student[stu].score[prob])
			student[stu].score[prob] = mark;
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= k; j++)
			if (student[i].score[j] != -1)
				student[i].total += student[i].score[j];
	sort(student + 1, student + n + 1, cmp);
	student[1].rank = 1;
	for (int i = 2; i <= n; i++)
	{
		if (student[i].total == student[i - 1].total)
			student[i].rank = student[i - 1].rank;
		else
			student[i].rank = i;
	}
	for (int i = 1; i <= n; i++)
		if (student[i].flag == true)
			output(student[i]);
	system("pause");
	return 0;
}