PTA PAT Judge 【模拟题，未完待续】

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, NNN (≤104\le 10^4≤10

​4

​​), the total number of users, KKK (≤5\le 5≤5), the total number of problems, and MMM (≤105\le 10^5≤10

​5

​​), the total number of submittions. It is then assumed that the user id’s are 5-digit numbers from 00001 to NNN, and the problem id’s are from 1 to KKK. The next line contains KKK positive integers p[i] (i=1, …, KKK), where p[i] corresponds to the full mark of the i-th problem. Then MMM 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 −1-1−1 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 -

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
using namespace std;
#define INF 0x3f3f3f
#define pi acos(-1.0)
#define MAX 100010
#define mod 9973
#define ll long long

/*

n, k,   m;
人 题目 提交

*/

const int N =1e5+10;
struct asd{
    int name;
    int sum;
    int a[10];
    bool vis[10];
    int ff;
};
asd q[N];
bool used[N];

int sco[10];int n,m,k;
bool cmp(asd z,asd x)
{
    if(z.sum>x.sum)
        return 1;
    else if(z.sum==x.sum)
    {
        int s1,s2;
        s1=s2=0;
        for(int i=1;i<=k;i++)
        {
            if(z.a[i]==sco[i])
                s1++;
            if(x.a[i]==sco[i])
                s2++;
        }
        if(s1>s2)
            return 1;
        else if(s1==s2)
        {
            if(z.name<x.name)
                return 1;
        }
    }
    return 0;
}

int main()
{

    memset(used,0,sizeof(used));
    cin>>n>>k>>m;
    for(int i=1;i<=k;i++)       //分数
        scanf("%d",&sco[i]);

    for(int i=1;i<=n;i++)       //初始化
    {
        q[i].name=i;            //名字
        for(int j=1;j<=5;j++)
            q[i].a[j]=0;        //分数都是0
        for(int j=1;j<=5;j++)
            q[i].vis[j]=0;      //提交
        q[i].sum=q[i].ff=0;     //sum分数，ff标记提交的那个
    }

    int mingzi,timu,fenshu;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&mingzi,&timu,&fenshu);
        used[mingzi]=1;                 //有这个人

        if(fenshu!=-1)                  //交了
        {
            q[mingzi].ff=1;         //就会输出的
            q[mingzi].vis[timu]=1;  //这道题目会输出；
        }

        if(fenshu>q[mingzi].a[timu])   //更新分数
        {
            q[mingzi].a[timu]=fenshu;
        }
    }

    for(int i=1;i<=n;i++)
    {
        int x=q[i].name;
        if(used[x])
        {
            for(int j=1;j<=k;j++)
            {
                q[x].sum+=q[x].a[j];
            }
        }
    }

    sort(q+1,q+1+n,cmp);

   /* for(int i=1;i<=n;i++)
    {
        int x=q[i].name;
        if(used[x])
        {
            printf("%d %d\n",q[i].name,q[i].sum);
        }
    }*/

    int tep,hh,xx;
    hh=xx=1;
    for(int i=1;i<=n;i++)
    {
        int x=q[i].name;
        if(used[x]&&q[i].ff)
        {
            printf("1");
            printf(" %05d %d",x,q[i].sum);
            for(int j=1;j<=k;j++)
            {
                if(q[i].vis[j])
                    printf(" %d",q[i].a[j]);
                else
                    printf(" -");
            }
            puts("");
            xx++;
            tep=q[x].sum;
            used[x]=0;
            break;
        }
    }

    for(int i=1;i<=n;i++)
    {
        int x=q[i].name;
        if(!used[x]||!q[i].ff)
            continue;
        if(tep>q[i].sum)
        {
            hh=xx;
            tep=q[i].sum;

            printf("%d %05d %d",hh,x,tep);

            for(int j=1;j<=k;j++)
            {
                if(q[i].vis[j])
                    printf(" %d",q[i].a[j]);
                else
                    printf(" -");
            }
            puts("");
        }
        else if(tep==q[i].sum)
        {
            printf("%d %05d %d",hh,x,tep);
            for(int j=1;j<=k;j++)
            {
                if(q[i].vis[j])
                    printf(" %d",q[i].a[j]);
                else
                    printf(" -");
            }
            puts("");
            xx++;
        }
    }
    return 0;
}

/*
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
*/