HDU 3697 Selecting courses（贪心）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3697

Problem Description

    A new Semester is coming and students are troubling for selecting courses. Students select their course on the web course system. There are n courses, the ith course is available during the time interval (A_i,B_i). That means, if you want to select the ith course, you must select it after time Ai and before time Bi. Ai and Bi are all in minutes. A student can only try to select a course every 5 minutes, but he can start trying at any time, and try as many times as he wants. For example, if you start trying to select courses at 5 minutes 21 seconds, then you can make other tries at 10 minutes 21 seconds, 15 minutes 21 seconds,20 minutes 21 seconds… and so on. A student can’t select more than one course at the same time. It may happen that no course is available when a student is making a try to select a course 

You are to find the maximum number of courses that a student can select.

 

Input

There are no more than 100 test cases.

The first line of each test case contains an integer N. N is the number of courses (0<N<=300)

Then N lines follows. Each line contains two integers Ai and Bi (0<=A_i<B_i<=1000), meaning that the ith course is available during the time interval (Ai,Bi).

The input ends by N = 0.

 

Output

For each test case output a line containing an integer indicating the maximum number of courses that a student can select.

 

Sample Input

2
1 10
4 5
0

 

Sample Output

2

 

Source

2010 Asia Fuzhou Regional Contest

题意：

有n门课程(n<=300),每门课程有个选课的时间段s,e,仅仅能在s之后。在e之前选择该门课程。

每位同学能够选择随意一个開始时间，然后每5分钟有一次选课机会，问每位同学最多能够选多少门课。

PS:

贪心。枚举開始的5个人时间！每次选择课程结束时间最早的。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 317;
struct sub
{
    int s;
    int e;
};
bool cmp(sub x, sub y)
{
    if(x.e == y.e)
    {
        return x.s < y.s;
    }
    return x.e < y.e;
}
int main()
{
    sub a[maxn];
    int t;
    int n;
    int vis[maxn];
    while(scanf("%d",&n) && n)
    {
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d",&a[i].s,&a[i].e);
        }
        sort(a,a+n,cmp);
        int ans = 0, cont = 0;
        for(int i = 0; i < 5; i++)
        {
            memset(vis,0,sizeof(vis));
            cont = 0;
            for(int j = i; j < a[n-1].e; j += 5)
            {
                for(int k = 0; k < n; k++)
                {
                    //printf("%d %d %d
",j,a[k].s,a[k].e);
                    if(!vis[k] && j>=a[k].s && j<a[k].e)
                    {
                        cont++;
                        vis[k] = 1;
                        break;
                    }
                }
            }
            if(ans < cont)
            {
                ans = cont;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}