PAT A1075 PAT Judge

PAT A1075 PAT Judge

题目描述：

　　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 -

参考代码：

/****************************************************
PAT A1075 PAT Judge
****************************************************/
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>

using namespace std;

struct stuExamInfo {
    int id = 0;
    int score[6] = { -1, -1, -1, -1, -1, -1 }; //成绩初始值设置为-1
    int totalScore = 0;
    int fullMrakCnt = 0;
    bool nonePasted = true;                    //是否没有有效的提交记录
};

bool myCmp(stuExamInfo a, stuExamInfo b) {
    if (a.nonePasted != b.nonePasted) return a.nonePasted < b.nonePasted;
    else if (a.totalScore != b.totalScore) return a.totalScore > b.totalScore;
    else if (a.fullMrakCnt != b.fullMrakCnt) return a.fullMrakCnt > b.fullMrakCnt;
    else return a.id < b.id;
}

int main() {
    int userCnt = 0, problemCnt = 0, recordCnt = 0;

    cin >> userCnt >> problemCnt >> recordCnt;

    //获取满分成绩
    vector<int> fullMrak(problemCnt, 0);
    for (int i = 0; i < problemCnt; ++i) {
        cin >> fullMrak[i];
    }

    //获取提交记录
    vector<stuExamInfo> userScoreList(userCnt);
    int userId = 0, problemId = 0, score = 0;
    for (int i = 0; i < recordCnt; ++i) {
        cin >> userId >> problemId >> score;

        //成绩不等于-1时说明是有效提交
        if (score != -1)
            userScoreList[userId - 1].nonePasted = false;
        //第一次提交不通过且对应题目无提交记录的时候成绩记为0
        if (score == -1 && userScoreList[userId - 1].score[problemId - 1] == -1) 
            score = 0; 
        //如果是第一次获得满分的提交，完美解决题目个数加1(注意之后提交的满分不能算数)
        if (score == fullMrak[problemId - 1] && userScoreList[userId - 1].score[problemId - 1] < score) 
            userScoreList[userId - 1].fullMrakCnt++;
        //如果提交的分数大于已有最高分，则更新成绩
        if (score > userScoreList[userId - 1].score[problemId - 1])
            userScoreList[userId - 1].score[problemId - 1] = score;
    }

    //计算总成绩
    for (int i = 0; i < userCnt; ++i) {
        userScoreList[i].id = i + 1;
        for (int j = 0; j < problemCnt; ++j) {
            if (userScoreList[i].score[j] != -1) {
                userScoreList[i].totalScore += userScoreList[i].score[j];
            }
        }
    }

    //对结果进行排序
    sort(userScoreList.begin(), userScoreList.end(), myCmp);

    //输出结果
    int rank = 1, lastTotalScore = userScoreList[0].totalScore;
    for (int i = 0; i < userCnt && userScoreList[i].nonePasted == false; ++i) {
        if (i != 0) cout << endl;

        if (userScoreList[i].totalScore != lastTotalScore) {
            rank = i + 1;
            lastTotalScore = userScoreList[i].totalScore;
        }
        cout << rank << ' ';
        cout << setw(5) << setfill('0') << userScoreList[i].id << ' ' << userScoreList[i].totalScore;
        for (int j = 0; j < problemCnt; ++j) {
            if (userScoreList[i].score[j] == -1) cout << ' ' << '-';
            else cout << ' ' << userScoreList[i].score[j];
        }
    }
    
    return 0;
}

注意事项：

　　1：注意有过一次满分的提交之后，再有该题目的满分提交时完美提交的题目数目是不会增加的。