2010 Asia Fuzhou Regional Contest 之 Selecting courses

Selecting courses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 2622    Accepted Submission(s): 717

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
   
   
   
   

贪心+暴力

把每门课按结束时间从小到大排序，结束时间相同按开始时间从小到大排序，这样按顺序选一定是最优的。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int vis[500];
struct road{
    int s,e;
}a[500];
int cmp(const road x,const road y){
    if(x.e==y.e)    return x.s<y.s;
    return x.e<y.e;
}
int main()
{
    int n;
    while(~scanf("%d",&n)){
        if(n==0)    break;
        for(int i=0;i<n;i++)
            scanf("%d%d",&a[i].s,&a[i].e);
        sort(a,a+n,cmp);
        int ans=0;
        for(int k=0;k<5;k++){
            int temp=0;
            memset(vis,0,sizeof(vis));
            for(int i=k;i<=a[n-1].e;i+=5){
                for(int j=0;j<n;j++){
                    if(vis[j])  continue;
                    if( i >= a[j].s && i < a[j].e){
                        vis[j]=1;
                        temp++;
                        break;
                    }
                }
            }
            ans=max(ans,temp);
        }
        printf("%d\n",ans);
    }
    return 0;
}