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 -. 1 #include <bits / STDC ++ H.> 2 the using namespace STD; . 3 struct the Result { . 4 int ID = 0 , Score [ . 6 ] = {- 2 , - 2 , - 2 , - 2 , - 2 , - 2 }, = rank . 1 , NUM = 0 , the Totalscore = 0 ; // number title id, each question scores, rankings, out of the total score . 5 BOOL in flag = to false ; // if there flag by compiled code, i.e., whether or not to output . 6 }; . 7 BOOL CMP ( const Result&r1,const Result&r2){//比较函数 8 if(r1.totalScore!=r2.totalScore) 9 return r1.totalScore>r2.totalScore; 10 else if(r1.num!=r2.num) 11 return r1.num>r2.num; 12 else 13 return r1.id<r2.id; 14 } 15 Result m[(int)(1e5+5)];//Result的数组 16 int main(){ 17 int N,K,M; 18 scanf("%d%d%d",&N,&K,&M); 19 int P[K+1];//存储每题的满分 20 for(int i=1;i<=K;++i) 21 scanf("%d",&P[i]); 22 while(M--){ 23 int a,b,c; 24 scanf("%d%d%d",&a,&b,&c); 25 m[a].id=a; 26 if(c>-1)// There are compiled code by 27 m [A] = .flag to true ; // set flag is to true 28 the else IF (C == - . 1 ) // not compile 29 C = 0 ; // set score of 0 30 m [A] .score [B] = max (m [A] .score [B], C); // update the score of the highest points title 31 is } 32 for ( int I = . 1 ; I <= N; + I +) // iterate Result 33 is IF (m [I] .flag) // required output 34 is for ( int J =. 1 ; J <= K; J ++) { // all traverse the examination subject 35 IF (m [I] .score [J] == P [J]) // had taken out of the subject 36 ++ m [ I] .num; // increment the number of out of Title 37 [ m [I] .totalScore = m + [I] .score [J] < 0 ? 0 : m [I] .score [J]; // update score 38 } 39 Sort (m + . 1 , m + N + . 1 , CMP); // Sort 40 for ( int I = 2 ; I <= N; I ++) // derived rank 41 is m [I] .rank m = [I ] .totalScore! = m [I- . 1?] .totalScore I: m [I- . 1 ] .rank; 42 is for ( int I = . 1 ; I <= N; I ++) // iterate Result 43 is IF (m [I] .flag) { // output 44 is the printf ( " % D% D% 05d " , m [I] .rank, m [I] .id, m [I] .totalScore); 45 for ( int J = . 1 ; J <= K; ++ J) 46 is IF (m [I] .score [J] < 0 ) // this problem or no submitted compile 47 the printf ( " - " ); // output - 48 else 49 printf(" %d",m[i].score[j]); 50 printf("\n"); 51 } 52 return 0; 53 }