PAT 1075. PAT Judge (25)(pat排名)（待修改）

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1075. PAT Judge (25)

时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 (<=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.

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

• 1.输入时如果提交为通过编译则分数记为-1。
• 2.总分数相同排名相同，再比较完全正确的题目个数，再比较id。
• 3.输出时，如果没提交过就记作-，提交过但未通过编译记作0.
• 4.如果没有提交任何能通过编译的，不用显示出来，意思是通过编译了但是没拿分的也要shown出来。

解题思路

• 1.要计算总分数，每道题的分数，每道题是否提交，有没有题通过编译，完全正确题目个数，以及算出排名。
• 2.把录入数据和求完全正确题目个数以及总分分开可能思路要清晰些。

未AC代码（最后一个没有通过）

#include<iostream>#include<vector>#include<cstdio>#include<algorithm>using namespace std;int n, k, m;struct student{  int id, t, rank,p;  bool h;  int visited[5];  int problem[5];  student() :id(-1), t(0), rank(1), p(0), h(false){}};//排序bool cmp(const student &a, const student &b){  if (a.t != b.t){    return a.t > b.t;  }  else if (a.p != b.p){    return a.p > b.p;  }  else{    return a.id < b.id;  }}student stu[10001];int fullMark[5];int main(){  scanf("%d %d %d", &n, &k, &m);  for (int i = 0; i < k; i++){    scanf("%d", &fullMark[i]);  }  //输入  int tem_id,prob_id,grade;  for (int i = 0; i < m; i++){    scanf("%d %d %d", &tem_id, &prob_id, &grade);    int tem_i = i;    i = tem_id;    stu[i].id = tem_id;    //用这个记录每个问题是否提交过，因为提交过没编译过记作0,否则要记作-    if (stu[i].visited[prob_id-1] == 0){      stu[i].visited[prob_id-1] = 1;    }    //用h记是否有题目通过    if (!stu[i].h&&grade>=0){      stu[i].h = true;    }    //取最大分数，注意初始值为0,所以输入为-1时不会更新，不会打扰到计算总分    if (stu[i].problem[prob_id - 1] < grade){      if (grade == fullMark[prob_id - 1]){        stu[i].p++;      }      stu[i].t = stu[i].t + grade - stu[i].problem[prob_id - 1];      stu[i].problem[prob_id - 1] = grade;    }    i = tem_i;  }  //排序  sort(stu + 1, stu + 1 + n, cmp);  int i = 1;  for (; i < n; i++){    if (stu[i + 1].h == false){      break;    }    if (stu[i + 1].t == stu[i].t){      stu[i + 1].rank = stu[i].rank;    }    else{      stu[i + 1].rank = i + 1;    }  }  //其实下面这个可以合并到上面那一步  for (int j = 0; j < i; j++){    int l = j + 1;    printf("%d %05d %d",stu[l].rank,stu[l].id,stu[l].t);    for (int d = 0; d < k; d++){      if (stu[l].visited[d] == 0){        printf(" -");      }      else{        printf(" %d", stu[l].problem[d]);      }    }    printf("\n");  }  return 0;}