Hdoj 3697 Selecting courses 贪心

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Selecting courses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 2082    Accepted Submission(s): 543


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 (Ai,Bi). 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<=Ai<Bi<=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,选课时有以下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。

先依照结束时间从小到大排序,由于是每过五分钟才干够选。那么我们仅仅须要枚举前四个时间,看是否在课程的起始与结束时间之内。就能够了。

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;
} 


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