【PAT】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 (<=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 sametotal_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 -

分析：（1）“-1”代表是没有通过编译，得分为0；没有提交该题则输出“-”则代表  （2）需要在排名中剔除的是：没有提交的；有提交但是没有能通过编译的；（3）得0分的需要统计
陷阱：输出的排名要注意，相同成绩的排名并列。如例子中的1 2 2 2 5。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#define MAX 6
struct node{
	vector<int> problem;  
	int total;
	int perfect;
	int id;	 
	bool submit;
};

bool cmp(node a, node b){
	if(a.total != b.total){
		return a.total > b.total;
	}else{
	 	if(a.perfect != b.perfect){
	 		return a.perfect > b.perfect;
		}else{
			return a.id < b.id;
		}
	}
}

int main(int argc, char** argv) {
	int userNum, proNum, subNum;
	scanf("%d%d%d",&userNum, &proNum, &subNum);
	vector<int> scores(proNum+1);
	int i, s;
	for(i=1; i<=proNum; i++){
		scanf("%d", &scores[i]);
	}
	
	int stuId, proId, sNow;
	vector<node> stus(userNum+1);
	for(i=1; i<=userNum; i++){
		stus[i].id= i;	 
		stus[i].problem.assign(MAX,-2);//-2代表这道题没有提交过 
		stus[i].total = -1;   
		stus[i].perfect = 0;
		stus[i].submit = false;
	}
	
	for(i=0; i<subNum; i++){
		scanf("%d%d%d", &stuId, &proId, &s);
		sNow = stus[stuId].problem[proId];
		if(!stus[stuId].submit){ 
			stus[stuId].submit = true; //标记为有提交过 
		}
		if(s>=0 && stus[stuId].total==-1){
			stus[stuId].total = 0;
		}	
		if(s>sNow){
			if(s==-1){
				stus[stuId].problem[proId] = 0;
			}else{
				stus[stuId].problem[proId] = s;
			}		
			if(s >= 0){
				//有得分，总得分要变化 
				if(sNow <= 0){					 
					stus[stuId].total += s;					
				}else{				 
					stus[stuId].total += (s-sNow);
				}
			} 							
			if(s == scores[proId]){
				stus[stuId].perfect ++;
			}
		} 
	}
	
	sort(stus.begin()+1, stus.end(), cmp);
	int rank = 1;
	int cnt = 0;
	for(i=1; i<stus.size(); i++){ 
		// submit=false 代表没有提交过
		// total=-1 代表没有编译通过的
		// total=0 && submit=true 代表有编译通过但是得分都为0的 
		cnt ++;
		if(stus[i].total==-1 || !stus[i].submit )
		   break;
		   
		if(i!=1){
			if(stus[i].total == stus[i-1].total){
				printf("%d",rank);
			}else{
				printf("%d",cnt);
				rank = cnt;
			}
		}else{
			printf("1");
		}
		printf(" %05d %d", stus[i].id,stus[i].total);
		for(int j=1; j<=proNum; j++){
			if(stus[i].problem[j] == -2){
				printf(" -");
			}else{
				if(stus[i].problem[j]!=-1)
				  printf(" %d",stus[i].problem[j]);
				else
				  printf(" 0");
			}		 
		}
		printf("\n");
	}	
	return 0;
}




另写了一个较为清晰的版本：

#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
#define proNum 6

struct node{
	node(){
		perfect = 0;
		total = 0;
		bool isShow = false;
	}
	bool isShow;
	int id;	
	int perfect;
	int total;
	map<int,int> m;
};

bool cmp(node a, node b){
	if(a.total!=b.total){
		return a.total > b.total;
	}else if(a.perfect != b.perfect){
		return a.perfect > b.perfect;
	}else {
		return a.id<b.id;
	}	
}

int N,K,M;
vector<int>scores;
vector<node>stu;
int main(int argc, char** argv) {
	int i;
	scanf("%d%d%d",&N,&K,&M);
	scores.resize(K+1);
	for(i=1; i<=K; i++){
		scanf("%d",&scores[i]);
	}
	int userId, proId, score;
	stu.resize(N+1);
	for(i=0; i<M; i++){
		scanf("%d%d%d",&userId,&proId,&score);
		if(score==-1){
			score = 0;	
		}else{
			if(!stu[userId].isShow){
				stu[userId].isShow = true;
			}
		}		
		stu[userId].id = userId;		
		map<int,int>::iterator it = stu[userId].m.find(proId);
		if(it==stu[userId].m.end()){
			stu[userId].total += score;
			stu[userId].m.insert(make_pair(proId,score));
			if(score == scores[proId])
				stu[userId].perfect++;
		}else if(it->second < score){
			stu[userId].total += score - it->second; 
			it->second = score;
			if(score == scores[proId])
			  stu[userId].perfect++;
		}		
	}
	sort(stu.begin(), stu.end(), cmp);
	int rank=1, cnt=0;
	int lastScore = stu[0].total;
	for(i=0; i<stu.size(); i++){
		node n = stu[i];
		if(!n.isShow){
			continue;
		}
		cnt++;
		if(n.total!=lastScore){
			rank = cnt;	
			lastScore = n.total;
		}
		printf("%d %05d %d",rank,n.id,n.total);
		map<int, int>::iterator it;
		for(int j=1; j<=K; j++){
			it = n.m.find(j);
			if(it==n.m.end()){
				printf(" -");
			}else{
				printf(" %d",it->second);
			}
		}
		printf("\n");
	}	
	return 0;
}