1075 PAT Judge (25分)(坑多，填坑)

最新推荐文章于 2021-03-25 12:54:15 发布

zbchenchanghao

最新推荐文章于 2021-03-25 12:54:15 发布

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分类专栏： PAT

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本文详细解析了PAT竞赛中排名系统的实现过程，包括输入规格、输出规格及样例输入输出。通过对用户提交的代码进行评分，生成基于得分、完美解决题目数量及用户ID的综合排名。文章分享了实现过程中遇到的问题及解决方案。

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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 −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 -

思路：这个题目还算常规，但是坑比较多，本人最后一个节点莽了2小时，终于搞定，原因：要求未提交或者提交了且没通过编译的人不参与排名。我直接所有的人都参与排名，有个标记决定此人是不是输出，但是会出现不输出的人排名在可输出的人之前，所以排名就G了（在总分默认为0的情况下，都没有一题满分，就肯定按id排序啊，例如一个人没有提交id为3，另一个人提交了且通过编译了，但是得了0分，此人id为6，那么排序的话那个不应该输出的排名在可输出的人之前，就错了），更改默认总分为-1，就可以了～

#include<iostream>
#include<vector>
#include<unordered_set>
#include<algorithm>
using namespace std;
struct node{
    int total_sore = -1; //坑
    int id;
    unordered_set<int> cnt;
    int pro_sore[6] = {0};
    bool visit[6] = {0}; 
    bool flag = false;
};
bool cmp(node a, node b){
    if(a.total_sore != b.total_sore){
        return a.total_sore > b.total_sore;
    }else
    {
        return a.cnt.size() != b.cnt.size() ? a.cnt.size() > b.cnt.size() : a.id < b.id;
    }
    
}
int main(){
    int n, k, m;
    
    scanf("%d %d %d", &n, &k, &m);
    vector<node> vec(n + 1);
    int full_sore[6] = {0};
    for(int i = 1; i <= k; i++){

        scanf("%d", &full_sore[i]);
    }
    for(int i = 0; i < m; i++){
        int id, num, sore;
        scanf("%d %d %d", &id, &num, &sore);
        vec[id].id = id;
        if(sore >= 0){
            vec[id].flag = true;
            if(vec[id].total_sore == -1){
                vec[id].total_sore = 0;
            }
        }
        if(vec[id].pro_sore[num] <= sore){
            
            vec[id].total_sore += (sore - vec[id].pro_sore[num]);
            vec[id].pro_sore[num] = sore;
            
            
        }
        vec[id].visit[num] = true;
        if(sore == full_sore[num]){
            vec[id].cnt.insert(num);
        }
        

    }
    sort(vec.begin() + 1, vec.end(), cmp);
    int rank = 1;
    for(int i = 1; i <= n; i++){
        
        if(vec[i].flag){
            if(i == 1 || vec[i].total_sore != vec[i - 1].total_sore){
                printf("%d ", i);
                rank = i;
            }else
            {
                printf("%d ", rank);
            }
            
            printf("%05d %d", vec[i].id, vec[i].total_sore);
            for(int j = 1; j <= k; j++){
                if(vec[i].visit[j]){
                    printf(" %d", vec[i].pro_sore[j]);
                }else
                {
                    printf(" -");
                }
                
                
            }
            printf("\n");
        }
        
    }

    return 0;
}