【PAT】1075. PAT Judge (25)【结构体的使用】

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题目描述

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.

翻译：PAT的排名是根据显示提交的得分情况的状态表进行排名的，这一次你需要得出PAT的排名表。

INPUT FORMAT

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.

翻译：每个输入文件包含一组测试数据。对于每组输入数据，第一行包括3个正整数N(<=10^4)，用户总数，K(<=5)，题目总数，和M(<=10^5)，提交总数。假设用户id是一个从00001到N的5位数字，问题的编号为从1到K。第二行包括K个正整数p[i] (i=1,…k)，p[i]代表第i题的满分分数。接着M行，每行按照以下格式给出一个提交记录：
user_id problem_id partial_score_obtained
如果提交不能通过编译器，则得分为-1，否则为一个[0,p[problem_id]]的整数。一行内所有数字之间用空格隔开。

OUTPUT FORMAT

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.

翻译：对于每组测试数据，你需要按照以下格式输出：
rank user_id total_score s[1] … s[K]
rank根据总分排名，所有拥有相同分数的用户获得同样的排名；s[i]代表第i题的部分得分。如果用户有一题没有提交过结果，则需要在对应位置输出“-”。如果一个用户一道题提交了多次，则输出最高的分数。
排名表必须按照排名升序输出。对于那些排名相同的人，用户必须按照完美解决题目的个数进行降序排序。如果仍然相同，则他们必须根据他们的id号进行升序输出。对于那些提交过题目未通过编译或从未提交过题目的人，他们将不会在排名中显示。数据保证至少有一个用户可以被展示在排名表上。

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 -

解题思路

模拟题目要求即可，只要有一道题得分>=0，就参与排名。根据总分->满分题目个数->ID号大小的方式排序。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#define INF 99999999
using namespace std;
int N,K,M; 
int score[10];
struct Stu{int id,sum,sco[10],flag;Stu(){sum=0;for(int i=0;i<10;i++)sco[i]=-2;flag=0;}int getSum(){for(int i=1;i<=K;i++){if(sco[i]>0)sum+=sco[i];if(sco[i]>=0&&!flag)flag++;if(sco[i]==score[i])flag++;}return flag;}bool operator<(const Stu &a)const{return sum==a.sum?(flag==a.flag?id<a.id:flag>a.flag):sum>a.sum;}
};
Stu stu[10010];
vector<Stu> ans;
int main(){scanf("%d%d%d",&N,&K,&M);for(int i=1;i<=K;i++)scanf("%d",&score[i]);int Tid,Tnum,Tscore; for(int i=0;i<M;i++){scanf("%d%d%d",&Tid,&Tnum,&Tscore);stu[Tid].id=Tid;stu[Tid].sco[Tnum]=max(stu[Tid].sco[Tnum],Tscore);}for(int i=1;i<=N;i++){if(stu[i].getSum())ans.push_back(stu[i]);}sort(ans.begin(),ans.end());int Grade=0,Score=-1;for(int i=0;i<ans.size();i++){if(ans[i].sum!=Score)Grade=i+1,Score=ans[i].sum;printf("%d %05d %d",Grade,ans[i].id,ans[i].sum);for(int j=1;j<=K;j++){if(ans[i].sco[j]>0)printf(" %d",ans[i].sco[j]);else if(ans[i].sco[j]>=-1)printf(" 0",ans[i].sco[j]);else printf(" -");}printf("\n");}return 0;
}