hdu 3697(贪心+部分枚举）

Selecting courses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 1300    Accepted Submission(s): 314

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

21 104 50

 

Sample Output

2

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 310;struct lesson{int s;int e;}l[maxn];int N , ans , L , R;bool cmp(lesson l1 , lesson l2){return l1.e < l2.e;}void readcase(){L = 10000;R = 0;ans = 0;for(int i = 0;i < N;i++){//cin >> l[i].s >> l[i].e;scanf("%d%d" , &l[i].s , &l[i].e);l[i].e--;L = min(L , l[i].s);R = max(R , l[i].e);}}void computing(){sort(l , l+N , cmp);//cout << L << " " << R << endl;for(int j = 0;j < 5;j++){int i = L+j;int vis[310] = {0};int temp = 0;while(i <= R){for(int k = 0;k < N;k++){if(vis[k] == 0 && l[k].s <= i && l[k].e >= i){vis[k] = 1;temp++;break;}}i += 5;}ans = max(ans , temp);//cout << ans <<" "<<L<< endl;if(ans >= N){break;}}//cout << ans << endl;printf("%d\n" , ans);}int main(){while(cin >> N &&N){readcase();computing();}return 0;}