PatA1075：PAT Judge

1、题目

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.

2、输入格式

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.

3、输出格式

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.

4、输入样例

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

 5、输出样例

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 -

6、代码

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct Student{

	int id, score[6];

	int total;    //总分     
        int pnum;	//满分题的个数

	bool show;    //是否输出，后续赋初始值false

}stu[10010];

bool compare(Student a, Student b){

	if(a.total != b.total) return a.total > b.total;    //成绩降序

	else if(a.pnum != b.pnum) return a.pnum > b.pnum;    //成绩相同，按满分题个数排名

	else return a.id < b.id;    //如果还相同，按id排名

}

int main(){

	int n, k, m, i, j;
       
	scanf("%d %d %d", &n, &k, &m);

	int problem_score[6];	//题目的分数
        
	for(i = 1; i <= k; i++) scanf("%d", &problem_score[i]);

	int xuhao, problem, score;
        
	for(i = 1; i <= n; i++){

		stu[i].id = i;    //把i的值赋给它，

		stu[i].show = false;
		
		stu[i].total = 0;

		stu[i].pnum = 0;

		for(j = 1; j <= k; j++) stu[i].score[j] = -2;	//给各题分数赋初始值

		//memset(stu[i].score, -2, sizeof(stu[i].score));	//这个不知道为什么不能用

	}
	
	for(i = 1; i <= m; i++){

		scanf("%d %d %d", &xuhao, &problem, &score);

		if(score == -1) score = 0;    //-1要求输出0
		
		if(stu[xuhao].score[problem] < score){
			
			stu[xuhao].score[problem] = score;	//初始值是-2，保留高分

			stu[xuhao].show = true;    //表示有数据，可以输出

		}

	}
	
	for(i = 1; i <= n; i++){

		for(j = 1; j <= k; j++){

			if(stu[i].score[j] >= 0) stu[i].total += stu[i].score[j];    //求总分

			if(stu[i].score[j] == problem_score[j]) stu[i].pnum++;	//统计满分的个数

		}

	}

	sort(stu + 1, stu + n + 1, compare);

	int rank = 1;
	
	for(i = 1; i <= n && stu[i].show; i++){

		if(i > 1 && stu[i].total != stu[i - 1].total) rank = i;	//分数相同，排名一样
	
		printf("%d %05d %d", rank, stu[i].id, stu[i].total);    //注意id的输出，用0补足5位

		for(j = 1; j <= k; j++){
			
			if(stu[i].score[j] != -2) printf(" %d", stu[i].score[j]);	//输出分数

			else printf(" -");    //题目没写就输出-
    
		}

		printf("\n");    //换行
		
	}

	return 0;

}
	

 7、备注

题目不难，前提是得读懂题目。唉，英语真难。