HDOJ3697 Selecting courses[贪心]

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：给n门课的选课开始时间和截止时间，你可以从任意时间开始，每隔五分钟选一门课，问最多能选多少课。只能在间隔五分钟的那个时间点选课，如果选不了就只能等到下一个五分钟。（刚好在课程开放时间边界也是不能选的）

这相当与是用一堆间隔为5的叉子去叉，看能叉中多少个。只需从0-5枚举开始时间即可。给的时间都是整数，边界又不能选，可以从0.5到4.5来枚举，防止边界判断有误。

就按结束时间排个序，找的时候就找第一门在时间范围内且没被选的就好。

#include<iostream>
#include<cassert>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<string>
#include<iterator>
#include<cstdlib>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define debug(x) cout<<"debug "<<x<<endl;
#define rep(i,f,t) for(int i = (f),_end_=(t); i <= _end_; ++i)
#define rep2(i,f,t) for(int i = (f),_end_=(t); i < _end_; ++i)
#define dep(i,f,t) for(int i = (f),_end_=(t); i >= _end_; --i)
#define dep2(i,f,t) for(int i = (f),_end_=(t); i > _end_; --i)
#define clr(c, x) memset(c, x, sizeof(c) )
typedef long long int64;
const int INF = 0x5f5f5f5f;
const double eps = 1e-8;


//*****************************************************

int n;
#define F first
#define S second
typedef pair<int,int> P;
P a[330];
bool vis[330];
int maxt;

int fn(double t){
    rep(i,1,n){
        if(vis[i])continue;
        if(a[i].F < t && t < a[i].S)return i;
    }
    return 0;
}
bool cmp(const P &p1,const P &p2){return p1.S < p2.S; }
int xuan(double tm){
	clr(vis,0);
	sort(a+1,a+n+1,cmp);
	int ans = 0;
	for(int i; tm < maxt; tm += 5){
		if( i = fn(tm) ){
			vis[i] = 1;
			++ans;
		}
	}
	return ans;
}
int main()
{
    a[0].S = INF;
    while(scanf("%d",&n), n){
    	maxt = 0;
    	rep(i,1,n){
    		scanf("%d%d",&a[i].F,&a[i].S);
    		if(a[i].S > maxt)maxt = a[i].S;
    	}
    	int ans = 0;
    	for(double i = 0.5; i < 5; i += 0.5){
    		int tmp = xuan(i);
    		ans = max(ans,tmp);
    	}
    	printf("%d
",ans);
    }
    return 0;
}

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