PTA_2019春_105_ PAT Judge

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

 这题比较简单，注释没有写。

用了冒泡，超时，又写了归并。

注意一下，这里面的指针数组！！！

/*PAT Judge*/
#include<stdio.h>
#include<stdlib.h>
struct personnode {
	int total;  /*总分*/
	int num;   /*满分个数*/
	int grade[6];  /*每题分数*/
	int flag;/*标记该学生有没有资格排序*/
	int data;  /*编号*/
};
typedef struct personnode* person;
void swap(person* a, person* b) {
	person temp;
	temp = *a;
	*a = *b;
	*b = temp;
}
void bubblesort(person* data, int N) {
	int i;
	for (int p = N - 1; p >= 0; p--) {
		int flag = 0;
		for (i = 0; i < p; i++) {
			if (data[i]->total <data[i + 1]->total) {
				swap(&data[i], &data[i + 1]);
				flag = 1;
			}
			if (data[i]->total == data[i + 1]->total) {
				if (data[i]->num < data[i + 1]->num) {
					swap(&data[i], &data[i + 1]);
					flag = 1;
				}
				if (data[i]->num == data[i + 1]->num) {
					if (data[i]->data > data[i]->data) {
						swap(&data[i], &data[i + 1]);
						flag = 1;
					}
				}
			}
		}
		if (flag == 0) break;
	}
}
void merge1(person* data, person* tempdata, int L, int R, int rightend) {
	int leftend, temp, numelements;
	int i;
	leftend = R - 1;
	temp = L;
	while (L <= leftend && R <= rightend) {
		if (data[L]->total >data[R]->total) {
			tempdata[temp++] = data[L++];
		}
		
		else if(data[L]->total==data[R]->total){
			if(data[L]->num>data[R]->num){
				tempdata[temp++] = data[L++];
			}
			else if(data[L]->num==data[R]->num){
				if(data[L]->data<data[R]->data){
					tempdata[temp++] = data[L++];
				}
				else{
					tempdata[temp++] = data[R++];
				}
			}
			else{
				tempdata[temp++] = data[R++];
			}
		}
		else  tempdata[temp++] = data[R++];
		/*else{
			tempdata[temp++] = data[R++];
		}*/
	}
	while (L <= leftend) {
		tempdata[temp++] = data[L++];
	}
	while (R <= rightend) {
		tempdata[temp++] = data[R++];
	}
}
void merge_pass(person* data, person* tempdata, int N, int length) {
	int i, j;
	for (i = 0; i < N - 2 * length; i += 2 * length) {
		merge1(data, tempdata, i, i + length, i + 2 * length - 1);
	}
	if (i + length < N) {
		merge1(data, tempdata, i, i + length, N - 1);
	}
	else {
		for (j = i; j < N; j++) {
			tempdata[j] = data[j];
		}
	}
}
void mergesort(person* data, int N) {   /*非递归*/
	int length = 1;
	person* tempdata = (person*)malloc(N * sizeof(struct personnode));
	while (length < N) {
		merge_pass(data, tempdata, N, length);
		length *= 2;
		merge_pass(tempdata, data, N, length);
		length *= 2;
	}
	free(tempdata);
}

int main() {
	int N, K, M;
	scanf("%d %d %d", &N, &K, &M);
	int* p = (int*)malloc((K + 1) * sizeof(int));
	for (int i = 1; i <= K; i++) {
		scanf("%d", &p[i]);
	}
	person data = (person)malloc((N + 1) * sizeof(struct personnode));
	for (int i = 1; i <= N; i++) {
		data[i].total = 0;
		data[i].num = 0;
		data[i].flag = 1;/*初始假设都有资格参与排序*/
		for (int j = 1; j <= 5; j++) {
			data[i].grade[j] = -2;/*对于每个题目初始化为-2*/
		}
	}
	for (int i = 1; i <= M; i++) {   /*total不能在这里计算，比较麻烦,但是total有一个作用，可以标记这个人有没有通过编译的*/
		int id,num,grade;
		scanf("%d %d %d", &id,&num,&grade);
		if (data[id].grade[num] < grade) {  /*如果这次输入的分数比之前的好则更新*/
			data[id].grade[num] = grade;   
			if (grade == p[num])  data[id].num++;
		}
	}
	/*计算总分*/
	for (int i = 1; i <= N; i++) {
		data[i].data = i;     /*注意一下编号在这里赋值*/
		int j = 0;
		for (int k = 1; k <= K; k++) {
			if (data[i].grade[k] < 0) j++;   /*没有参加，或者没有编译通过的个数*/
			if (data[i].grade[k] > 0) {
				data[i].total += data[i].grade[k];
			}
			
		}
		if (j == K) data[i].flag = 0;
	}
	/*
	for (int i = 1; i <= N; i++) {
		printf("%05d %d %d %d ", i,data[i].total,data[i].num,data[i].flag);
		for (int k = 1; k <= K; k++) {
			printf("%d ", data[i].grade[k]);
		}
		printf("\n");
	}
	printf("----------\n");*/
	/*排序*/
	/*间接排序，归并排序*/
	person* A = (person*)malloc((N)* sizeof(person));/*指针数组*/
	for (int i = 0; i < N; i++) {
		A[i] = data + i + 1;  /*每一个指针指向一个struct personnode的地址*/
	}
	/*
	for (int i = 0; i < N; i++) {
		printf("%05d %d %d %d ", i+1, A[i]->total, A[i]->num, A[i]->flag);
		for (int k = 1; k <= K; k++) {
			printf("%d ", A[i]->grade[k]);
		}
		printf("\n");
	}
	printf("----------\n");
	*/
	mergesort(A, N);
	//bubblesort(A, N);
	int m = 0 ;
	for (int i = 0; i < N; i++) {
		if(A[i]->flag==0) continue;
		if(i>0&&(A[i]->total==A[i-1]->total)){
			printf("%d ",m);
		}
		else{
			printf("%d ",i+1);
			m = i+1;
		}	
		printf("%05d %d ",A[i]->data, A[i]->total);
		for (int k = 1; k <= K; k++) {
			if(A[i]->grade[k] ==-1){
				printf("0");
			}
			else if(A[i]->grade[k]==-2){
				printf("-");
			}
			else{
				printf("%d", A[i]->grade[k]);
			}
			if(k!=K) printf(" ");
		}
		printf("\n");
	}
//	printf("----------\n");

	return 0;
}