1075 PAT Judge (25 分)-CFANZ编程社区

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 (≤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 -

本题不难，但是卡了好久，主要题目没有读懂。。。英语太渣。。。题意好几处理解错了

唯一编程难点:录入时要建立hash表，有点麻烦

编译错误0分不是-1分

又是题意理解错误英文太差了
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
未提交或者提交了一次编译都未过的不输出也即是编译过了但是总分为0的还是要排名

排名升序-》单题最高分降序-》id升序

编译不过记0 从未提交记-1

#include<bits/stdc++.h>
using namespace std;
struct User{
  int id;//%05d输出 
  int grade[6],sumGrd,Index,perfectNum;//各题最高成绩  总分  总排名  满分题数量 
  bool passed;//编译通过过一次 
  User(){
    for(int i=1;i<=5;i++) grade[i]=-2;
    perfectNum=0;
    sumGrd=0;
    passed=false;
  }
}user[10100];
bool compare(User a,User b){
  if(a.sumGrd!=b.sumGrd) return a.sumGrd>b.sumGrd;
  if(a.perfectNum!=b.perfectNum) return a.perfectNum>b.perfectNum;
  if(a.passed!=b.passed) return a.passed>b.passed;//true1>false0
  return a.id<b.id;
} 

int N,K,M,p[10];//人数  问题数   提交数   分值数组 
int main(){
  //freopen("in.txt","r",stdin);
  cin>>N>>K>>M;
  for(int i=1;i<=K;i++) cin>>p[i];
  
  //录入数据 
  int n,score,id,Count=0,index;//临时题号  临时分数  临时id  Count当前人数   index临时序号 
  int HashId[10100];
  memset(HashId,-1,sizeof(HashId)); 
  for(int i=0;i<M;i++){//i<M 不是N千万别弄错了   输入很复杂   
    cin>>id>>n>>score;
    if(HashId[id]==-1) HashId[id]=Count++;
    index=HashId[id];
    user[index].id=id;
    if(score>=0) user[index].passed=true; 
    if(user[index].grade[n]<score) user[index].grade[n]=score;
    if(user[index].grade[n]==-1) user[index].grade[n]=0; //接着这么处理吧 但是要加个标记 
  }
  
  //计算总分  每个人的       顺便计算满分题数量 
  for(int i=0;i<N;i++){
    for(int j=1;j<=K;j++){
      if(user[i].grade[j]>0){//未提交的负数只是标记肯定不能加 
        if(user[i].grade[j]==p[j]) user[i].perfectNum++; 
        user[i].sumGrd+=user[i].grade[j];
      }
    }
  }
  
  sort(user,user+N,compare);
  //计算排名
  for(int i=0;i<N;i++){
    if(user[i].passed==false) break;
    user[i].Index=i+1;
    if(i>0&&user[i].sumGrd==user[i-1].sumGrd) user[i].Index=user[i-1].Index;
  } 
  
  for(int i=0;i<N;i++){
    if(user[i].passed==false) break;
    printf("%d %05d %d",user[i].Index,user[i].id,user[i].sumGrd); 
    for(int j=1;j<=K;j++) {
      if(user[i].grade[j]<0) cout<<" -";
      else cout<<" "<<user[i].grade[j];
    }
    cout<<endl;
  }
  return 0;
}