HDU 3697 Selecting courses（贪心+暴力）（2010 Asia Fuzhou Regional Contest）

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.

 

题目大意：给n个课程，每个课程有一个选择时间段(ai, bi)，只能在这个时间段里面选择。然后你可以在任何时刻开始选课，然后每隔5分钟可以选一门课，问最多选多少门课。

思路：首先开始是越早越好，在8分开始不如在3分就开始。然后枚举5个开始时间，然后对每个时间枚举可以选择的课，选bi最小的（因为bi最小的被后面的时间选中的可能性最小，若bi能在后面被选中那么其他也能在后面被选中）。完全暴力不需要任何优化即可AC。

PS：那个(ai, bi)应该是开区间，不过我们可以选择在0分1秒的时候开始，所以可以看成是[ai, bi)。

 

代码（31MS）：

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <cstring>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 const int MAXN = 310;

 8 

 9 int a[MAXN], b[MAXN];

10 bool sel[MAXN];

11 int n;

12 

13 int solve() {

14     int ret = 0;

15     for(int st = 0; st < 5; ++st) {

16         memset(sel, 0, sizeof(sel));

17         int ans = 0;

18         for(int i = st; i <= 1000; i += 5) {

19             int best = -1;

20             for(int j = 0; j < n; ++j)

21                 if(!sel[j] && a[j] <= i && i < b[j]) {

22                     if(best == -1) best = j;

23                     else if(b[j] < b[best]) best = j;

24                 }

25             if(best != -1) {

26                 sel[best] = true;

27                 ++ans;

28             }

29         }

30         if(ret < ans) ret = ans;

31     }

32     return ret;

33 }

34 

35 int main() {

36     while(scanf("%d", &n) != EOF && n) {

37         for(int i = 0; i < n; ++i) scanf("%d%d", &a[i], &b[i]);

38         printf("%d\n", solve());

39     }

40 }

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