hdu 3697 Selecting courses (暴力+贪心)

Selecting courses

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

Total Submission(s): 1856    Accepted Submission(s): 469

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

 

题意是：每次选一个起始点，以后仅仅能在+5这些点上进行选课。由于数据范围比較小。能够暴力。

枚举0、1、2、3、  4这5个起点，对于每一个确定的起点能够得到一些能够进行选课的点，把选课时间段以终点进行从小到大排序，枚举可得最优解。另外，区间（a,b）能够变为[a,b)，由于对于a分钟来说。a分01秒是能够选的。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 1005
const int inf=0x3fffffff;
int mark[N];
struct node
{
    int l,r;
}g[305];
bool cmp(node a,node b)
{
    return a.r<b.r;
}
int main()
{
    int i,j,k,r,n;
    while(scanf("%d",&n),n)
    {
        r=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&g[i].l,&g[i].r);
            r=max(r,g[i].r);
        }
        sort(g,g+n,cmp);
        int ans=0;
        for(i=0;i<5;i++)
        {
            memset(mark,0,sizeof(mark));
            for(j=i;j<r;j+=5)
                mark[j]=1;
            int tmp=0;
            for(j=0;j<n;j++)
            {
                for(k=g[j].l;k<g[j].r;k++)
                {
                    if(mark[k])
                    {
                        mark[k]=0;
                        tmp++;
                        break;
                    }
                }
            }
            ans=max(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}