1075 PAT Judge (25分)+测试点分析

问题

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 (≤10​4​​), the total number of users, K (≤5), the total number of problems, and M (≤10​5​​), 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 -

解决方法

分析：排序；排序的顺序是总分大>满分通过的数目多>学生id小。这道题模仿PAT的得分情况对其排序进行输出。

#include
#include
#include
#include
using namespace std;
struct student
{
     
	int cnt = 0;
	int id,rank;
	int fullpass;
	int totalscore;
	int getscore[6] = {
     -2,-2,-2,-2,-2,-2};
};
bool cmp(student a, student b)
{
     
	if (a.totalscore != b.totalscore) return a.totalscore > b.totalscore;
	else if (a.fullpass != b.fullpass) return a.fullpass > b.fullpass;
	else return a.id < b.id;
}
int main()
{
     
	int n, k, m;
	scanf("%d %d %d\n", &n, &k, &m);
	int fullscore[6] = {
      0};
	vector<struct student>vec(n + 1);
	for (int i = 1; i <=k; i++)	scanf("%d", &fullscore[i]);
	for (int i = 0; i < m; i++)
	{
     
		int id,question, getscore;
		scanf("%d %d %d", &id, &question, &getscore);
		if (getscore>=0&&getscore > vec[id].getscore[question])
		{
     
			if (getscore == fullscore[question]&&vec[id].getscore[question]<fullscore[question]) vec[id].fullpass++;
			vec[id].getscore[question] = getscore;
			vec[id].cnt++;
		}
		else if (getscore == -1 && vec[id].getscore[question] == -2) vec[id].getscore[question] = -1;
	}
	for (int i = 0; i < vec.size(); i++)
	{
     
		for (int j = 0; j < 6; j++)
		{
     
			if(vec[i].getscore[j]>=0) vec[i].totalscore += vec[i].getscore[j];
		}
		vec[i].id = i;
	}
	sort(vec.begin(), vec.end(), cmp);
	vec[0].rank = 1;
	for (int i = 1; i<vec.size(); i++)
	{
     
		if ( vec[i - 1].totalscore == vec[i].totalscore) vec[i].rank = vec[i-1].rank;
		else vec[i].rank = i + 1;
	}
	for (int i = 0; i < vec.size(); i++)
	{
     
		if (vec[i].cnt)
		{
     
			printf("%d %05d %d", vec[i].rank, vec[i].id, vec[i].totalscore);
			for (int j = 1; j <= k; j++)
			{
     
				if (vec[i].getscore[j] == -2) printf(" -");
				else if (vec[i].getscore[j] == -1) printf(" 0");
				else printf(" %d", vec[i].getscore[j]);
			}
			printf("\n");
		}
	}
	return 0;
}

测试点分析

这道题按照答题思路就可以得到19分，只有测试点2和最后一个测试点过不去。测试点二是测试提交通过实得0分和提交编译不通过定义为0分的区分。
测试点4是测试（1）满分重复通过，对满分重复通过的计数会不会重复累加；
（2）前面提交问题K有得分，但是之后又提交了一次-1分。所以对于分数是-1的情形，需要先判断该题是否曾经有得分，然后再决定是否要更新。

后记

题的难度不大，难以得满分的原因在于：测试点2和最后一个测试点的通过。