PAT_A1075#PAT Judge

Source:

PAT A1075 PAT Judge (25 分)

Description:

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 (≤), the total number of users, K (≤), the total number of problems, and M (≤), 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_scoreobtain 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 -

Keys:

  • 模拟题

Code:

 1 /*
 2 Data: 2019-07-14 19:26:17
 3 Problem: PAT_A1075#PAT Judge
 4 AC: 43:25
 5 
 6 题目大意:
 7 排名
 8 输入:
 9 第一行给出,考生数N<=1e4(考号1~N),问题数K<=5(题号1~K),提交数M<=1e5
10 接下来一行,各题分值
11 接下来M行,user_id,pro_id,score(-1表示未通过编译)
12 输出:
13 排序按照,总分,满分数,id(各题均为提交,或均未通过编译,视为无效)
14 次序,user_id,total_socre,ith_score(未提交‘-’,未通过编译‘0’)
15 */
16 #include<cstdio>
17 #include<vector>
18 #include<string>
19 #include<iostream>
20 #include<algorithm>
21 using namespace std;
22 const int M=1e4+10,K=10;
23 struct node
24 {
25     int id,score,full;
26     int pro[K],valid;
27 }info[M];
28 int mark[K];
29 vector<node> ans;
30 
31 bool cmp(const node &a, const node &b)
32 {
33     if(a.score != b.score)
34         return a.score > b.score;
35     else if(a.full != b.full)
36         return a.full > b.full;
37     else
38         return a.id < b.id;
39 }
40 
41 int main()
42 {
43 #ifdef    ONLINE_JUDGE
44 #else
45     freopen("Test.txt", "r", stdin);
46 #endif
47 
48     int n,m,k;
49     scanf("%d%d%d", &n,&k,&m);
50     for(int i=1; i<=k; i++)
51         scanf("%d", &mark[i]);
52     for(int i=1; i<=n; i++)
53     {
54         info[i].id = i;
55         info[i].valid=info[i].full=info[i].score=0;
56         for(int j=1; j<=k; j++)
57             info[i].pro[j]=-2;
58     }
59     for(int i=0; i<m; i++)
60     {
61         int id,grade,index;
62         scanf("%d%d%d", &id,&index,&grade);
63         if(info[id].pro[index] < grade)
64         {
65             if(grade != -1)
66                 info[id].valid=1;
67             else
68                 grade = 0;
69             info[id].pro[index] = grade;
70             if(grade == mark[index])
71                 info[id].full++;
72         }
73     }
74     for(int i=1; i<=n; i++)
75         if(info[i].valid == 1)
76         {
77             for(int j=1; j<=k; j++)
78                 if(info[i].pro[j]!=-2)
79                     info[i].score += info[i].pro[j];
80             ans.push_back(info[i]);
81         }
82     sort(ans.begin(),ans.end(),cmp);
83     int r=1;
84     for(int i=0; i<ans.size(); i++)
85     {
86         if(i!=0 && ans[i-1].score!=ans[i].score)
87             r=i+1;
88         printf("%d %05d %d", r,ans[i].id,ans[i].score);
89         for(int j=1; j<=k; j++)
90             if(ans[i].pro[j]==-2)   printf(" -");
91             else printf(" %d",ans[i].pro[j]);
92         printf("\n");
93     }
94 
95     return 0;
96 }

 

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转载自www.cnblogs.com/blue-lin/p/11185499.html
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