[天坑]1075 PAT Judge （25 分)

2019-04-04

[天坑]1075 PAT Judge （25 分)

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 (≤104), the total number of users, K (≤5), the total number of problems, and M (≤105), 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:

1	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:

1	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 -

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出pat的提交记录，要求各个用户的排名。注意：
1.按总分高至低，总分相同者名次相同。
2.总分相同时，按满分题数排序。
3.满分题数相同时，按照id低至高排序。
4.没有提交记录者，或者都未通过编译者，不计入排序

分析

用结构体user记录用户。其中有id，各题分数，满分题数，总分，是否提交、通过编译。各题初始化为-2，易于判断是未提交或未通过编译。读入各用户信息，然后排序即可。但是光这样还不够，有个地方非常隐蔽。由于编译未通过或未提交都是0分，但不输出。提交得0分是要输出的。都为0分时，按id排序，未通过编译者可能排在0分者前，导致名次错误。对应的是测试点2，这个地方卡了好久。因此，总分初始化为-1，有成功提交记录再初始化为0。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;

struct user{
    int id;
    vector<int> scr;
    int pfts;
    int tal;
    bool submit;
    bool comp;

    user() {this->tal=-1;}
};
bool cmp(user u1,user u2){
    return u1.tal!=u2.tal?u1.tal>u2.tal:(u1.pfts!=u2.pfts?u1.pfts>u2.pfts:u1.id<u2.id);
}

int main() {
    int n,k,m;
    cin>>n>>k>>m;
    vector<user> users(n+1);

    int mark[k+1];
    for(int i=1;i<=n;i++) {
        users[i].scr.resize(k + 1);
        //各题分数初始化为-2
        for(int j=1;j<=k;j++)
            users[i].scr[j]=-2;
        users[i].id=i;
    }
    for(int i=1;i<=k;i++)
        cin>>mark[i];
    int uid,pid,scr;
    for(int i=0;i<m;i++){
        cin>>uid>>pid>>scr;
        //对应id有提交记录
        users[uid].submit=true;
        if(scr>=0) {
            //编译通过
            users[uid].comp = true;
            //总分初始化为0
            if(users[uid].tal==-1)
                users[uid].tal=0;
        }
        if(scr>users[uid].scr[pid]) {
            if(users[uid].scr[pid]>=0)//原来已有分数，总分增加相应增量
                users[uid].tal+=scr-users[uid].scr[pid];
            else if(scr>=0)//原来未提交或编译未通过，总分增加对应分数
                users[uid].tal+=scr;
            users[uid].scr[pid] = scr;
            if(scr==mark[pid])//该题满分
                users[uid].pfts++;
        }
    }
    sort(users.begin(),users.end(),cmp);

    int r=1;
    for(int i=0;i<n;i++){
        user u=users[i];
        if(u.submit&&u.comp) {
            printf("%d %05d %d", r, u.id, u.tal);
            for (int j = 1; j <= k; j++) {
                int i = u.scr[j];
                cout << " ";
                if(i==-2)
                    cout<<"-";
                else if(i==-1)
                    cout<<'0';
                else
                    cout<<i;
            }
            cout << endl;
            if (i < n-1 && u.tal != users[i + 1].tal)
                r=i+2;
        }
    }
    return 0;
}

其他

pat的系统也非常迷，第一次提交超时，查了半天没发现哪错了，再提交又过了。