enter 時間 hit size man style accepted 不能 ria

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 (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

題意：有n門課，選課時有以下rule：

1：每種課都有起始和結束，必須在之內選取。

2：每次選取之後5分鐘後不能再選課。

先依照結束時間從小到大排序，由於是每過五分鐘才幹夠選。那麽我們僅僅須要枚舉前四個時間，看是否在課程的起始與結束時間之內。就能夠了。

代碼：

#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;
} 

Hdoj 3697 Selecting courses 【貪心】