HDU-3697

Selecting courses

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 3138    Accepted Submission(s): 874

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

 

Source

2010 Asia Fuzhou Regional Contest

 

Recommend

chenyongfu

题意：输入若干线段，l r 表示在这两个数之间可以选课 不能在这两个端点的时刻选课，就是在给出的选课线段区间内 在每5分钟选一次课的情况下  求出做大选课数量是多少

贪心+枚举

1000ms ，解决方案是枚举0-4 因为 往后的数都是0-4分别加5的倍数得到的 0-4 分别表示在0.5 1.5 2.5 3.5 4.5 处选课 记为初始值 i

需要先对终点排序 由于 我们要先选早结束的放前面 把晚结束的往后放 这里贪心一下 每次选完标记线段 当前初始值i下 不要再选

我们枚举这5个数 在不断加5加到区间r的最大值 然后对每个数到线段里判断 如果在线段里而不是在端点 就++ 标记线段 下次直接跳过

这里枚举了一下

code:

#include<bits/stdc++.h>using namespace std;typedef long long ll;struct node{int l,r;}s[310];bool bok[310]; bool cmp(node a,node b){return a.r<b.r;}int main(){int t;ios::sync_with_stdio(0);while(cin>>t,t){int m=0,ans = -1;for(int i=1;i<=t;i++){cin>>s[i].l>>s[i].r;m = max(s[i].r,m);}sort(s+1,s+1+t,cmp);for(int i=0;i<5;i++){//0.5 1.5 2.5 3.5int c=0;for(int j=1;j<=t;j++)bok[j]=0;for(int k = i;k<m;k+=5) {   for(int p = 1;p<=t;p++){if(bok[p])continue;else if(k<s[p].r&&k>=s[p].l)//在区间段里 {c++;bok[p]=1;break; }}} ans = max(ans,c);} cout<<ans<<endl;}return 0;}