The ranklist of PAT is generated from the status list, which shows the scores of the submittions. 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 submittions. 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 submittion in the following format:
user_id problem_id partial_score_obtained
where partial_score_obtained is either -1 if the submittion 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 -
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=10010; struct Student{ int id; //准考证号 int score[6]; //每道题对应的得分(k<=5) int score_all; //总分 int solve; //完美解决的题数（得全分无bug） bool flag; //是否有能通过编译的提交 }stu[maxn]; int n, k, m; //n个考生，k道题，m次提交记录 int full[maxn]; //每道题的满分 //排序函数 bool cmp(Student a, Student b){ if(a.score_all!=b.score_all) return a.score_all>b.score_all; else if(a.solve!=b.solve) return a.solve>b.solve; else return a.id<b.id; } //初始化 void init() { for(int i=1; i<=n; i++){ //从下标1开始存放信息 stu[i].id=i; //准考证号记为i stu[i].score_all=0; //总分初始化为0 stu[i].solve=0; //完美解决的题数初始化为0 stu[i].flag=false; //false:没有能通过编译的提交 memset(stu[i].score, -1, sizeof(stu[i].score)); //某学生每道题目得分初始化为-1 //划重点：memset(a, 0, sizeof(a)): 对数组a[]赋初值 0 or -1 } } int main() { scanf("%d%d%d", &n, &k, &m); //n个考生，k道题，m次提交记录 //步骤(1): 初始化 init(); for(int i=1; i<=k; i++) scanf("%d", &full[i]); //full[i]: 每道题的满分 //步骤(2): 录入信息并按要求排序 int u_id, p_id, score_get; //考生id, 题目id, 所获分值（-1: 表示编译未通过） for(int i=0; i<m; i++) { scanf("%d%d%d", &u_id, &p_id, &score_get); //不是编译错误，则该考生有能通过编译的提交 if(score_get!=-1) { stu[u_id].flag=true; } //某题第一次编译错误，分值记为0分，便与输出 if(score_get==-1 && stu[u_id].score[p_id]==-1) { stu[u_id].score[p_id]=0; } //某题第一次获得满分，则完美解题数+1 if(score_get==full[p_id] && stu[u_id].score[p_id]<full[p_id]) { stu[u_id].solve++; } //某题获得更高分值，则覆盖 if(score_get>stu[u_id].score[p_id]) { stu[u_id].score[p_id]=score_get; } } for(int i=1; i<=n; i++){ //n个考生，k道题，m次提交记录 for(int j=1; j<=k; j++){ if(stu[i].score[j] != -1) //计算总分 stu[i].score_all += stu[i].score[j]; } } sort(stu+1, stu+1+n, cmp); //按要求排序 //步骤(3):按要求输出： 排名 准考证号 总分 k科成绩 int r=1; //当前排名 for(int i=1; i<=n && stu[i].flag==true; i++){ if(i>1 && stu[i].score_all != stu[i-1].score_all) r=i; //当前考生分数不同于前一位考生分数，则排名为在该考生之前的总考生数 printf("%d %05d %d", r, stu[i].id, stu[i].score_all); for(int j=1; j<=k; j++){ //输出该考生各科成绩 if(stu[i].score[j] == -1) printf(" -"); //没有提交过 else printf(" %d", stu[i].score[j]); } printf(" "); } return 0; }