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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 positive integers, N (≤10​4​​), the total number of users, K (≤5), the total number of problems, and M (≤10​5​​), 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 -

 題目大意：輸入包括n個學生，編號從00001到N，包括K個題目，m次提交，對它們進行排序：

1.一個學生多次對一個題目提交，要取分數最高的一次；

2.如果一個學生沒有提交過，或者是提交了均未通過編譯，也就是輸入未-1，那麼不計入統計中，注意：這裡提交了得分為0，和-1是不一樣的，前者是通過了編譯但是得分為0，後者是未通過編譯。

我的AC：

#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
using namespace std;

struct Stu{
    int id,total,perfect,pass;//pass記錄通過編譯的題目數，如果為0，那麼不進入list。
    int task[6];//先存上5道題的陣列。
    Stu(){
        fill(task,task+6,-1);
        total=0;perfect=0;pass=0;
    }
}stu[100000];
int score[6];
vector<Stu> vt;
bool cmp(Stu&a,Stu&b){
    if(a.total>b.total)return true;
    else if(a.total==b.total&&a.perfect>b.perfect)return true;
    else if(a.total==b.total&&a.perfect==b.perfect) return a.id<b.id;
    return false;
}
int main() {
    int n,k,m;
    cin>>n>>k>>m;
    for(int i=1;i<=k;i++){
        cin>>score[i];
    }
    int id,tid,sco;
    for(int i=0;i<m;i++){
        cin>>id>>tid>>sco;
        if(sco!=-1)stu[id].pass+=1;//此處只要!=0就可以計入排序。
        if(sco==-1)sco=0;
        stu[id].task[tid]=max(sco,stu[id].task[tid]);//分數每次都取最高的那個。
        stu[id].id=id;
        //提交多次的要取得分最高的那次。
//        if(sco!=-1)
//            stu[id].total+=sco;
//        if(sco==score[tid])//AC數目也不能在這裡判斷，有可能會多次AC
//            stu[id].perfect+=1;//記錄AC的題目數。
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=k;j++){
            if(stu[i].task[j]!=-1)
                stu[i].total+=stu[i].task[j];
            if(stu[i].task[j]==score[j])
                stu[i].perfect+=1;
        }
        if(stu[i].pass!=0){
            vt.push_back(stu[i]);
        }
    }
    sort(vt.begin(),vt.end(),cmp);
    int rank=1;
    printf("1 %05d %d",vt[0].id,vt[0].total);
    for(int i=1;i<=k;i++){
        if(vt[0].task[i]==-1)
           printf(" -");
        else
            printf(" %d",vt[0].task[i]);
    }
    printf("\n");
    for(int i=1;i<vt.size();i++){
        if(vt[i].total!=vt[i-1].total){
            rank=i+1;
        }
        printf("%d %05d %d",rank,vt[i].id,vt[i].total);
        for(int j=1;j<=k;j++){
            if(vt[i].task[j]==-1)
                printf(" -");
            else
                printf(" %d",vt[i].task[j]);
        }
        printf("\n");
    }
    return 0;
}

 

//總體來說就是根據題目的要求來寫，整個過程思路清晰的話，還是比較簡單的。