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贪心 HDU 3697

2015-01-27 06:24:48 · 作者: · 浏览: 8
标签: 贪心 HDU 3697

因为时间是都是5分钟增加的。所以我们可以从0开始到4每次都把可能的时间保存到一个数组里。然后遍历所有的情况

对于结构体的话，我们先按照结束的时间从小到大来排序，如果结束的时间相同，那么就按照开始的时间从小到大来排序

最后在判断的时间的时候就要一个闭区间一个开区间了。据说只要是时间判断的话都要这样的。

Selecting courses

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 1986 Accepted Submission(s): 504

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
Then N lines follows. Each line contains two integers Ai and Bi (0<=A i 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

Source 2010 Asia Fuzhou Regional Contest 
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                   using namespace std; #define MAXN 300 + 10 struct node{ int x; int y; }a[MAXN]; int vis[MAXN]; int t[1111]; bool cmp(node a,node b){ if(a.y == b.y){ return a.x < b.x; } return a.y < b.y; } int Max(int a,int b){ return a>b?a:b; } int main(){ int n,i,j; int time; while(~scanf("%d",&n)){ if(n == 0){ break; } for(i=0;i
                   
                     mark){ break; } t[cnt] = flag; cnt++; } sum = 0; for(i=0;i
                    
                     =a[j].x && t[i]