hdu 3697 贪心

Selecting courses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)

Total Submission(s): 1851    Accepted Submission(s): 468

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 

题目描述：有n个课程，每个课程有一个时间段选课（为开区间），一个人可以从任意时刻选课，每隔五分中选一次课，问最多能够选多少们课？

分析： 这里只给出解法，先将每个选课区间按结束时间排序，然后枚举前0--4分钟，之后枚举选课时间，对每个选课时间看能否选一门课注意这里只能选择排序后的最早满足条件的一门，之后记录最大值即可

#include<stdio.h>
#include<algorithm>
#include<string.h>
#define N 400
using namespace std;
int vist[N];
struct node
{
    int x,y;
}a[N];
bool cmp(node a,node b)
{
    if(a.y!=b.y)
        return a.y<b.y;
    else
        return a.x<b.x;
}
int main()
{
    int n,i,max,j,k,num;
    while(scanf("%d",&n),n!=0)
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        sort(a,a+n,cmp);
        max=-1;
        for(i=0;i<5;i++)
        {
            memset(vist,0,sizeof(vist));
            num=0;
            for(j=i;j<a[n-1].y;j+=5)
            {
                for(k=0;k<n;k++)
                {
                    if(!vist[k]&&a[k].x<=j&&j<a[k].y)
                    {
                        num++;
                        vist[k]=1;
                        break;
                    }
                }
            }
            if(num>max)
                max=num;
        }
        printf("%d\n",max);
    }
    return 0;
}