HDU 3697 Selecting courses（贪心）

最新推荐文章于 2022-11-13 20:15:43 发布

转载最新推荐文章于 2022-11-13 20:15:43 发布 · 56 阅读

文章标签：

#php

本文介绍了一个课程选择问题的解决方法，通过贪心算法实现学生在有限时间内选择最多课程的目标。问题涉及课程时间段的选择限制，算法通过枚举开始时间和优选最早结束时间的课程来最大化所选课程数量。

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\n",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\n",ans);
    }
    return 0;
}