首页 > ACM题库 > HDU-杭电 > hdu 2310 Analyzing Login/Logout Records-枚举[解题报告]C++
2014
01-05

hdu 2310 Analyzing Login/Logout Records-枚举[解题报告]C++

Analyzing Login/Logout Records

问题描述 :

You have a computer literacy course in your university. In the computer system, the login/logout records of all PCs in a day are stored in a file. Although students may use two or more PCs at a time, no one can log in to a PC which has been logged in by someone who has not logged out of that PC yet.You are asked to write a program that calculates the total time of a student that he/she used at least one PC in a given time period (probably in a laboratory class) based on the records in the file.

The following are example login/logout records.

* The student 1 logged in to the PC 1 at 12:55
* The student 2 logged in to the PC 4 at 13:00
* The student 1 logged in to the PC 2 at 13:10
* The student 1 logged out of the PC 2 at 13:20
* The student 1 logged in to the PC 3 at 13:30
* The student 1 logged out of the PC 1 at 13:40
* The student 1 logged out of the PC 3 at 13:45
* The student 1 logged in to the PC 1 at 14:20
* The student 2 logged out of the PC 4 at 14:30
* The student 1 logged out of the PC 1 at 14:40

For a query such as “Give usage of the student 1 between 13:00 and 14:30″, your program should answer “55 minutes”, that is, the sum of 45 minutes from 13:00 to 13:45 and 10 minutes from 14:20 to 14:30, as depicted in the following figure.

输入:

The input is a sequence of a number of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 10.Each dataset is formatted as follows.

N M
r
record1

recordr
q
query1

queryq
The numbers N and M in the first line are the numbers of PCs and the students, respectively. r is the number of records. q is the number of queries. These four are integers satisfying the following.

1 ≤ N ≤ 1000, 1 ≤ M ≤ 10000, 2 ≤ r ≤ 1000, 1 ≤ q ≤ 50

Each record consists of four integers, delimited by a space, as follows.

t n m s

s is 0 or 1. If s is 1, this line means that the student m logged in to the PC n at time t . If s is 0, it means that the student m logged out of the PC n at time t . The time is expressed as elapsed minutes from 0:00 of the day. t , n and m satisfy the following.

540 ≤ t ≤ 1260, 1 ≤ n ≤ N , 1 ≤ m ≤ M

You may assume the following about the records.
# Records are stored in ascending order of time t.
# No two records for the same PC has the same time t.
# No PCs are being logged in before the time of the first record nor after that of the last record in the file.
# Login and logout records for one PC appear alternatingly, and each of the login-logout record pairs is for the same student.

Each query consists of three integers delimited by a space, as follows.

ts te m

It represents “Usage of the student m between ts and te “. ts , te and m satisfy the following.

540 ≤ ts < te ≤ 1260, 1 ≤ m ≤ M

输出:

The input is a sequence of a number of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 10.Each dataset is formatted as follows.

N M
r
record1

recordr
q
query1

queryq
The numbers N and M in the first line are the numbers of PCs and the students, respectively. r is the number of records. q is the number of queries. These four are integers satisfying the following.

1 ≤ N ≤ 1000, 1 ≤ M ≤ 10000, 2 ≤ r ≤ 1000, 1 ≤ q ≤ 50

Each record consists of four integers, delimited by a space, as follows.

t n m s

s is 0 or 1. If s is 1, this line means that the student m logged in to the PC n at time t . If s is 0, it means that the student m logged out of the PC n at time t . The time is expressed as elapsed minutes from 0:00 of the day. t , n and m satisfy the following.

540 ≤ t ≤ 1260, 1 ≤ n ≤ N , 1 ≤ m ≤ M

You may assume the following about the records.
# Records are stored in ascending order of time t.
# No two records for the same PC has the same time t.
# No PCs are being logged in before the time of the first record nor after that of the last record in the file.
# Login and logout records for one PC appear alternatingly, and each of the login-logout record pairs is for the same student.

Each query consists of three integers delimited by a space, as follows.

ts te m

It represents “Usage of the student m between ts and te “. ts , te and m satisfy the following.

540 ≤ ts < te ≤ 1260, 1 ≤ m ≤ M

样例输入:

4 2
10
775 1 1 1
780 4 2 1
790 2 1 1
800 2 1 0
810 3 1 1
820 1 1 0
825 3 1 0
860 1 1 1
870 4 2 0
880 1 1 0
1
780 870 1
13 15
12
540 12 13 1
600 12 13 0
650 13 15 1
660 12 15 1
665 11 13 1
670 13 15 0
675 11 13 0
680 12 15 0
1000 11 14 1
1060 12 14 1
1060 11 14 0
1080 12 14 0
3
540 700 13
600 1000 15
1000 1200 11
1 1
2
600 1 1 1
700 1 1 0
5
540 600 1
550 650 1
610 620 1
650 750 1
700 800 1
0 0

样例输出:

55
70
30
0
0
50
10
50
0


题意:给出每个人在某台电脑的登录时间与下线时间,求在某个时间段某个人的上线时间。

题解:大模拟~可以先记录每个人的上下线时间点,看做一个区间的左右界限,然后记录区间深度,如果到某个时间点时,区间深度大于0,就说明这个人在线,否则,就不在。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
struct data
{
    int p,t;
    bool lg;
    data(){}
    data(int _p,int _t,bool _lg){p=_p;t=_t;lg=_lg;}
};
vector<data> po[10005];
int main()
{
    int n,m;
    //freopen("data.txt","r",stdin);
    while(scanf("%d%d",&n,&m),n||m)
    {
        for(int i=1;i<=m;i++)
            po[i].clear();
        int r,q,t,nn,mm,s;
        scanf("%d",&r);
        for(int i=0;i<r;i++)
        {
            scanf("%d%d%d%d",&t,&nn,&mm,&s);
            po[mm].push_back(data(nn,t,s));
        }
        scanf("%d",&q);
        for(int i=0;i<q;i++)
        {
            int x,y,len,ans=0,lx;
            scanf("%d%d%d",&x,&y,&mm);
            lx=x;
            len=po[mm].size();
            int cnt=0;
            for(int j=0;j<len;j++)
            {
                if(po[mm][j].t>y)
                {
                    if(cnt>0)
                        ans+=y-lx;
                    break;
                }
                else if(po[mm][j].t>=x)
                {
                    if(cnt>0)
                        ans+=po[mm][j].t-lx;
                    lx=po[mm][j].t;
                }
                if(po[mm][j].lg)
                    cnt++;
                else
                    cnt--;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

 


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