1075 PAT Judge (25 point(s))_1075 pat judge (25 point(s))-优快云博客

1075 PAT Judge (25 point(s))

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 -

常规模拟题，难度中档。主要是要理清思路，不要陷入混乱，尽可能减少代码量。

涉及点：1. 结构体的灵活使用（函数、重载运算符等）；2.名次赋值

此题因为限制了K范围不大于5，因此直接在结构体中设置那么多数组，不会占用太多内存，如果K太大可能导致MLE。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int N,K,M;//users pros submissions
struct user{
	int id;int rank;
	bool isSolved[6]={false};
	bool isCompiled[6]={false};
	bool hasSubmitted[6]={false};
	bool isShown=false;
	int marks[6];
	int solvedCnt=0;
	int totalScore=0;
	void shown(){
		for(int i=1;i<=K;i++){
			if(isCompiled[i]){
				isShown=true;break;
			}
		}
	}
	void scoreSum(){
		for(int i=1;i<=K;i++) totalScore+=marks[i];
	}
	void solvedSum(){
		for(int i=1;i<=K;i++){
			if(isSolved[i]) solvedCnt++;
		}
	}
	bool operator < (const user &other) const{
		if(totalScore!=other.totalScore) return totalScore>other.totalScore;
		if(solvedCnt!=other.solvedCnt) return solvedCnt>other.solvedCnt;
		return id<other.id;
	}
};
const int MAX = 1e4+7;
int full[6];
user U[MAX];
int main(void){
	cin>>N>>K>>M;
	for(int i=1;i<=N;i++){
		U[i].id = i;
	}
	for(int i=1;i<=K;i++) cin>>full[i];
	int id,proid,score;
	while(M--){
		cin>>id>>proid>>score;
		U[id].hasSubmitted[proid]=true;
		if(score>=0){
			U[id].isCompiled[proid] = true;
			if(score>U[id].marks[proid]){
				U[id].marks[proid]=score;
				if(score==full[proid]){
					 U[id].isSolved[proid] = true;
				}
			}
		}
	}
	for(int i=1;i<=N;i++){
		U[i].scoreSum();
		U[i].solvedSum();
		U[i].shown();
	}
	sort(U+1,U+N+1);
	for(int i=1;i<=N;i++){
		if(i==1||U[i].totalScore!=U[i-1].totalScore){
			U[i].rank=i;
		} else if(U[i].totalScore==U[i-1].totalScore){
			U[i].rank=U[i-1].rank;
		}
	}
	for(int i=1;i<=N;i++){
		if(U[i].isShown){
			printf("%d %05d %d",U[i].rank,U[i].id,U[i].totalScore);
			for(int j=1;j<=K;j++){
				if(U[i].hasSubmitted[j]==false ) cout<<" -";
				else cout<<" "<<U[i].marks[j];
			}
			cout<<endl;
		}
	}
	return 0;
}