PAT (Advanced Level) Practise 1075 PAT Judge (25)

1075. PAT Judge (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 2020 25 25 3000002 2 1200007 4 1700005 1 1900007 2 2500005 1 2000002 2 200005 1 1500001 1 1800004 3 2500002 2 2500005 3 2200006 4 -100001 2 1800002 1 2000004 1 1500002 4 1800001 3 400001 4 200005 2 -100004 2 0

Sample Output:

1 00002 63 20 25 - 182 00005 42 20 0 22 -2 00007 42 - 25 - 172 00001 42 18 18 4 25 00004 40 15 0 25 -

题意：有n个人，有k个题目，有m次提交记录，按每个人的得分进行排名，分数一样的名词相同，输出时，分数一样的按完美过题数来排名（一题都没有编译成功的或没提交的不输出）

解题思路：模拟，这题一个人一题完美过了也有可以再次提交这题，需要进行标记，而且编译成功了也可能零分

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;struct node{    int id,sum,cnt,Rank;    int x[8],vis[8];    friend bool operator <(node a,node b)    {        if(a.sum!=b.sum) return a.sum>b.sum;        else if(a.cnt!=b.cnt) return a.cnt>b.cnt;        else return a.id<b.id;    }}a[10009],ans[10009];int p[8],flag[10005];int main(){    int n,k,m;    while(~scanf("%d%d%d",&n,&k,&m))    {        memset(flag,0,sizeof flag);        for(int i=1;i<=n;i++)        {            a[i].id=i,a[i].sum=a[i].cnt=0;            memset(a[i].x,-1,sizeof a[i].x);            memset(a[i].vis,0,sizeof a[i].vis);        }        for(int i=1;i<=k;i++) scanf("%d",&p[i]);        while(m--)        {            int id1,id2,sorce;            scanf("%d%d%d",&id1,&id2,&sorce);            if(sorce>-1) flag[id1]=1;            a[id1].x[id2]=max(a[id1].x[id2],max(sorce,0));            if(sorce==p[id2]&&!a[id1].vis[id2])            {                a[id1].cnt++;                a[id1].vis[id2]=1;            }        }        for(int i=1;i<=n;i++)            for(int j=1;j<=k;j++)                if(a[i].x[j]!=-1) a[i].sum+=a[i].x[j];        int cnt=1;        for(int i=1;i<=n;i++)            if(flag[i]) ans[cnt++]=a[i];        sort(ans+1,ans+cnt);        ans[1].Rank=1;        for(int i=2;i<cnt;i++)        {            if(ans[i].sum==ans[i-1].sum) ans[i].Rank=ans[i-1].Rank;            else ans[i].Rank=i;        }        for(int i=1;i<cnt;i++)        {            printf("%d %05d %d",ans[i].Rank,ans[i].id,ans[i].sum);            for(int j=1;j<=k;j++)            {                if(ans[i].x[j]==-1) printf(" -");                else printf(" %d",ans[i].x[j]);            }            printf("\n");        }    }    return 0;}