前言

昨天PAT官网迁移服务器，所以就去牛客网刷题了，测试点打死都过不去。死磕到三点未果，今天PAT好了立刻去测了一下就过了，牛客网害人啊。。。

这是第一次开坑，之前刷了一部分的题但是都没有发文，所以做个题解报告合集方便后人参考使用吧。相关源码我都更新在gitee上了需要自取xingleigao/study - Gitee.com

题目描述

1075 PAT Judge (25 分)

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.

输入描述

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.

输出描述

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.

输入示例

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

结尾无空行

输出示例

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 -

结尾无空行

解题思路

题目理解

PAT A级别的所有题目都是英文，其实题主作为一个六级都没考过的人都能看懂，相信大家都能看懂吧，看不懂也没关系，经常出现的单词也不多，多看就好了.

一共有N个考生。K道题，M个提交记录。其中考生id为5位数字，第一行会给出N、K、M，第二行给出M道题的每题的分值。然后接下来M行给出提交记录。其中提交中分值-1表示编译不通过，其它表示分值。然后按要求排序。

1.按总分排序 2.总分相同按完美解决题数排序 3.前两项相同按id排序

输出规则：

1.总分相同则排名也相同

2.若考生无提交记录或者没有能编译通过的提交，则该考生的信息不输出

3.对需要输出的考生，若某题未通过编译，记0分；如果没有提交输出“-”。

代码实现

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct student{
    int id;         //准考证号
    int score[6];   //每道题分数
    bool flag;      //标记是否有通过编译的提交
    int score_sum;  //记录总分
    int solve;      //记录解决题目数
};

bool cmp(student a,student b){
    if(a.score_sum != b.score_sum)  return a.score_sum > b.score_sum;       //总分不同降序排序总分
    else if(a.solve != b.solve) return a.solve > b.solve;                   //总分相同降序排完美解决数字
    else return a.id < b.id;                                                //前两项相同升序id
}

void initstu(student* stu, int n){
    for(int i = 1; i <= n; ++i){
        stu[i].id = i; //设置结构体初始id
        memset(stu[i].score,-1,sizeof(stu[i].score));//分数初始为-1
    }  
}
int main(){
    int n,k,m,full[6];
    scanf("%d%d%d", &n,&k,&m);
    for(int i = 1;i <= k; ++i)   scanf("%d",&full[i]);//读入每题的分值

    struct student stu[n+1];    //创建结构体数组存储
    memset(stu, 0, sizeof(stu)); //初始所有元素为0
    initstu(stu,n);             //初始化结构体id
    int u_id,p_id,p_score;      //读入的临时变量
    for(int i = 0; i < m;++i){
        scanf("%d%d%d",&u_id,&p_id,&p_score);//读入数据
        //if(u_id > n || p_id > k || p_id < 1) continue;
        if(p_score != -1)   stu[u_id].flag = true;//有编译通过的数据则记录
        if(p_score == -1 && stu[u_id].score[p_id] == -1)   stu[u_id].score[p_id] = 0;   //有提交记录则标记为1方便统计
        if(p_score == full[p_id] && stu[u_id].score[p_id] != p_score)   stu[u_id].solve ++;//第一次完美解决做统计
        if(p_score > stu[u_id].score[p_id]) stu[u_id].score[p_id] = p_score;    //更高分数覆盖

    }
    for(int i = 1;i <= n;++i){  //统计总分
        int sum = 0;            //临时变量保存避免频繁访存
        for(int j = 1; j <= k; ++j)
            if(stu[i].score[j] != -1)   sum += stu[i].score[j];
        stu[i].score_sum = sum;
    }

    sort(stu + 1, stu + n +1 ,cmp); //排序数组
    int paiming = 1;
    for(int i = 1; i <= n && stu[i].flag;++i){//格式输出
        if(i > 1 && stu[i].score_sum != stu[i - 1].score_sum)    paiming = i;
        printf("%d %05d %d",paiming, stu[i].id, stu[i].score_sum);//输出排名、id以及总分
        for(int j = 1; j <= k; j++){
            if(stu[i].score[j] == -1)   printf(" -");//没有提交记录
            else printf(" %d",stu[i].score[j]);
        }
        printf("\n");
    }

    return 0;
}