PAT 甲级 1075 PAT Judge

1075 PAT Judge （25 point(s)）

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 -

经验总结：

这一题，就是看起来十分复杂的一道排序题，先说说里面需要注意的地方：
1. 当然就是排序规则，首先，要排除那些没有一题编译通过的同学，然后再对剩余的同学先按照总分降序，再按照完美通过的题数降序，再按照学号升序。
2. 注意，这里，如果一个人提交的分数是-1，代表编译未通过，但是如果这个人其他题目有通过的，说明他会被输出，而提交了未通过的题目不应该输出 ‘ - ’ 而是应该输出 ‘ 0 ’ ，这算是这一题比较隐蔽的一个坑点。
3. 排序方面，我是先记录所有合法的同学数，然后先排一次，把合法的全部排到了不合法的前面，然后只对合法的进行后续排序，至于排名就很简单啦，就不多说了~

AC代码

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int INF=0x3fffffff;
const int maxn=10010;
int n,k,m,p[6];
struct node
{
	int uid,quest[6],pernum,total,rank;
	bool valid;
	node()
	{
		valid=false;
		pernum=0;
		memset(quest,-1,sizeof(quest));
		total=0;
	}
}Node[maxn];
bool cmp(node a,node b)
{
	return a.valid>b.valid;
}
bool cmp1(node a,node b)
{
	if(a.total!=b.total)
		return a.total>b.total;
	if(a.pernum!=b.pernum)
		return a.pernum>b.pernum;
	return a.uid<b.uid;
}
int main()
{
	int no,score,pid,num=0;
	scanf("%d%d%d",&n,&k,&m);
	for(int i=1;i<=k;++i)
		scanf("%d",&p[i]);
	for(int i=0;i<m;++i)
	{
		scanf("%d%d%d",&no,&pid,&score);
		Node[no].uid=no;
		if(score!=-1&&Node[no].valid==false)
		{
			Node[no].valid=true;
			++num;
		}
		if(score==-1&&Node[no].quest[pid]==-1)
			Node[no].quest[pid]=0;
		if(score>Node[no].quest[pid])
		{
			if(Node[no].quest[pid]!=-1)
				Node[no].total-=Node[no].quest[pid];
			Node[no].total+=score;
			Node[no].quest[pid]=score;
			if(score==p[pid])
				Node[no].pernum++;
		}
	}
	sort(Node,Node+maxn,cmp);
	sort(Node,Node+num,cmp1);
	Node[0].rank=1;
	for(int i=1;i<num;++i)
	{
		Node[i].rank=i+1;
		if(Node[i].total==Node[i-1].total)
			Node[i].rank=Node[i-1].rank;
	}
	for(int i=0;i<num;++i)
	{
		printf("%d %05d %d",Node[i].rank,Node[i].uid,Node[i].total);
		for(int j=1;j<=k;++j)
		{
			if(Node[i].quest[j]==-1)
				printf(" -");
			else
				printf(" %d",Node[i].quest[j]);
		}
		printf("\n");
	}
	return 0;
}