Selecting courses

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

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：有n门课，选课时有下面rule：

1：每种课都有起始和结束，必须在之内选取。

2：每次选取之后5分钟后不能再选课。

先按照结束时间从小到大排序，因为是每过五分钟才可以选，那么我们只需要枚举前四个时间，看是否在课程的起始与结束时间之内，就可以了。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
const int M = 1100;
using namespace std;

struct node{
	int l, r;
}s[M];
int n;
bool vis[M];

bool cmp(node a, node b){
	return a.r < b.r;
}

int f(double x){
	memset(vis, 0, sizeof(vis));
	int res = 0;
	for(double d = x; d < M; d += 5){
		for(int i = 0; i < n; ++ i){
			if(!vis[i]&&d > s[i].l &&s[i].r > d){
				res++;
				vis[i] = 1; break;
			}
		}
	}
	return res;
}

int main(){
	while(scanf("%d", &n), n){
		for(int i = 0; i < n; ++ i) scanf("%d%d", &s[i].l, &s[i].r);
		sort(s, s+n, cmp);
		int ans = 0;
		for(double i = 0.5; i < 5; ++ i){
			ans = max(ans, f(i));
		}
		printf("%d\n", ans);
	}
	return 0;
}