1075 PAT Judge (25)
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 (<=10^4^), the total number of users, K (<=5), the total number of problems, and M (<=10^5^), 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 -
题目大意：有n个学生， k道题， m条提交， 给出每一题的满分。
有m条提交信息，格式为：user_id problem_id partial_score_obtained。partial_score_obtained，表示所得分数，为-1的时候表示编译没有通过
输出要求：按照分数从高到低排序， 分数相同的具有一样的名次。 当总分一样的时候按照得满分个数降序排序， 如果依然一样，则按照id升序排序。 如果某个人没有提交任何解答，或者提交的都没有通过编译，则不输出这个人的信息；如果某个人的某一体没有提交过，则在对应的位置输出'-1';
思路：建立一个student类，记录id， 满分解答的个数(perfect), 记录每一题的得分(score[]), 记录每一题是否提交（submit[]）, 记录有无有效提交(flag, 有提交，并且至少有一个通过编译)， 记录总分(total);
通过input()函数对输入进行处理，如果提交的分数比之前的高，则更新对应的分数，总分，改变标志位。
注意点：当提交没有通过编译的时候，默认的是得分为0， 这就会和提交通过了且得分为0重合，需要单独判断。
最后一个点没有通过测试，找不出原因
经过不断的尝试，发现当有多个人的成绩为0的时候，名次会出现错误； 出现错误原因是，当分数为0的时候，排序是按照id来排序的，导致在确定名次的时候出现错误。这样的错误不容易发现。原因在于自己的数据结构构造的不是很好。在对成绩初始化的时候，不应该初始化为0.因为提交了答案没有通过编译，该题的得分是0，未提交该题，该题得分也是0，还有提交了得分为0的情况。这里有三种状态，却用一种方式来表达，导致这种不容易发现的错误。以后的编程中，对于不同的状态应用不同的值来表示，才能避免这种不容易察觉的错误
这次的修改没有改变数据结构，而是改动的比较函数，当分数为0的时候，flag为true的优先级更高，就解决了上面的问题。
检查错误的方法还有待提高
1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 using namespace std; 5 6 class node{ 7 public: 8 int id, total, perfect; 9 int score[6], submit[6]; 10 bool flag; 11 node(){total=0; perfect=0; flag=false; id=0; for(int i=0; i<6; i++){score[i]=0; submit[i]=0;}} 12 }; 13 vector<node> v; 14 vector<int> p(6); 15 void input(int id, int num, int score){ 16 if(score>v[id].score[num]){ 17 v[id].total = v[id].total+score-v[id].score[num]; 18 v[id].score[num]=score; 19 v[id].submit[num]=1; 20 v[id].flag = true; 21 if(score==p[num]) v[id].perfect++; 22 }else if(score==0){ 23 v[id].flag=true; 24 v[id].submit[num]=1; 25 } 26 if(score==-1) v[id].submit[num]=1; 27 } 28 29 bool cmp(node a, node b){ 30 if(a.total != b.total) return a.total>b.total; 31 if(a.perfect != b.perfect) return a.perfect > b.perfect; 32 if(a.total==0 && b.total==0) return a.flag-b.flag>0; 33 return a.id<b.id; 34 } 35 int main(){ 36 int n, k, m, i; 37 cin>>n>>k>>m; 38 v.resize(n+1); 39 for(i=1; i<=k; i++) cin>>p[i]; 40 for(i=0; i<m; i++){ 41 int id, num, score; 42 scanf("%d %d %d", &id, &num, &score); 43 v[id].id=id; 44 input(id, num, score); 45 } 46 sort(v.begin()+1, v.end(), cmp); 47 int rank=1; 48 for(i=1; i<v.size(); i++){ 49 if(v[i].flag){ 50 if(i==1) printf("1 "); 51 else{ 52 if(v[i].total==v[i-1].total) printf("%d ", rank); 53 else{printf("%d ", i); rank=i;} 54 } 55 printf("%05d %d", v[i].id, v[i].total); 56 for(int j=1; j<=k; j++){ 57 if(v[i].submit[j]==1) printf(" %d", v[i].score[j]); 58 else printf(" -"); 59 } 60 printf("\n"); 61 } 62 } 63 return 0; 64 }
