PAT Judge

PAT Judge

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 − 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 -

 

解题思路

　　这道题不难，但是细节特别多，需要好好审题，逻辑不要乱了！

#include <cstdio>
#include <iostream>
#include <algorithm>

struct User {
    int id, sum, pass, rank;
    int *score;
    bool flag;
};

bool cmp(User &u1, User &u2);

int main() {
    int n, k, m;
    scanf("%d %d %d", &n, &k, &m);
    User user[n + 1];
    int prob[k + 1];
    for (int i = 1; i <= n; i++) {
        user[i].score = new int[k + 1];
        std::fill(user[i].score, user[i].score + k + 1, -1);    // score数组初始化为-1表示这道题完全没有提交过 
        user[i].id = i;
        user[i].sum = 0;
        user[i].pass = 0;
        user[i].flag = false;
    }
    for (int i = 1; i <= k; i++) {
        scanf("%d", prob + i);
    }
    
    for (int i = 1; i <= m; i++) {
        int id, p, s;
        scanf("%d %d %d", &id, &p, &s);
        if (user[id].score[p] == -1 && s == -1) {   // 这道题没有提交过并且第一次提交时编译不通过 
            user[id].score[p] = 0;                  // 分数改为0分 
        }
        else {                                      // 否则题目提交过，或者没提交过并且第一次提交通过编译 
            if (s > user[id].score[p]) {            // 如果得分比之前要高，更新。否则小于或等于都不更新 
                if (user[id].score[p] != -1) user[id].sum += s - user[id].score[p]; // 更新总分数 
                else user[id].sum += s;
                user[id].score[p] = s;              // 更新该题的分数 
                if (s == prob[p]) user[id].pass++;  // 同时，如果该题得了满分，通过次数加1 
            }
            user[id].flag = true;                   // 有题目通过编译，标识为true，可以打印成绩 
        }
    }
    
    std::sort(user + 1, user + n + 1, cmp);
    
    for (int i = 1; i <= n; i++) {
        if (user[i].flag) { // 有题目通过编译才可以打印成绩 
            if (i == 1 || user[i - 1].sum != user[i].sum) user[i].rank = i;
            else user[i].rank = user[i - 1].rank;
            printf("%d %05d %d", user[i].rank, user[i].id, user[i].sum);
            for (int j = 1; j <= k; j++) {
                if (user[i].score[j] == -1) printf(" -");
                else printf(" %d", user[i].score[j]);
            }
            putchar('\n');
        }
    }
    
    return 0;
}

bool cmp(User &u1, User &u2) {
    // 优先级：总分 > 题目通过数 > 考生编码 
    if (u1.sum != u2.sum) return u1.sum > u2.sum;
    else if (u1.pass != u2.pass) return u1.pass > u2.pass;
    else return u1.id < u2.id;
}