1075 PAT Judge（最后一个测试点详解。。）

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 (≤10​ ), the total number of users, K (≤5), the total number of problems, and M (≤10​), 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 -

神坑！！哎，被坑哭了。（也是自己的不仔细的原因）
题目要求：
按照总分，通过数，ID进行排序，最后一个测试点有两个坑：提交重复的满分和先提交有分数之后又提交一次编译不通过就会覆盖原成绩。其实在接收数据的时候就存储就可以很好的解决两个问题。
我被自己蠢哭的点是等级不能在判断是否提交且通过编译器的模块中！！！因为会有所有题目都没通过的提交但它的sum为0，就导致通过编译但sum为0的user可能排在它的后面。

#include
#include
#include
using namespace std;
struct usr{
	vector<int> score;
	int sum = 0, id, rank, perfect = 0;
	bool submit = false;//是否提交且通过编译器
}; 
bool cmp(const usr &x, const usr &y){
	if(x.sum != y.sum) return x.sum > y.sum;
	else if(x.perfect != y.perfect) return x.perfect > y.perfect;
	else return x.id < y.id;
}	
int main(){ 
	int n, k, m, maxs[6];
	scanf("%d %d %d", &n, &k, &m);
	vector<usr> u(n + 1); 
	for(int i = 1; i <= n; i++) u[i].score.resize(k + 1, -1);//每题分数默认为-1
	for(int i = 1; i <= k; i++) scanf("%d", &maxs[i]);
	for(int i = 0; i < m; i++){
		int uid, pid, score;
		scanf("%d %d %d", &uid, &pid, &score); 
		u[uid].id = uid;
		if(score >= 0)//如果大于等于0则可以输出
			u[uid].submit = true;
		else //如果score为-1则置0
			score = 0; 
		if(u[uid].score[pid] < score){//只有在score大于已存的分数时才操作
			if(u[uid].score[pid] > 0)
				u[uid].sum -= u[uid].score[pid];
			u[uid].score[pid] = score;
			u[uid].sum += score;
			if(score == maxs[pid])//不会遇到重复的最大值
				u[uid].perfect += 1;
		}
	}
	sort(u.begin() + 1, u.end(), cmp);
	for(int i = 1; i <= n; i++){
		if(i != 1 && u[i].sum == u[i - 1].sum)
                u[i].rank = u[i - 1].rank; 
        else
                u[i].rank = i;
        if(u[i].submit == true){
			printf("%d %05d %d", u[i].rank, u[i].id, u[i].sum);
			for(int j = 1; j <= k; j++){
				if(u[i].score[j] != -1)
					printf(" %d", u[i].score[j]);
				else
					printf(" -");
			}
			printf("\n");
		}
	}
	return 0;
}