[英]How to check if two or more time periods overlap each other in c++?
I am using std::pair to get start time and end time of each time periods and I put all those time periods in an array (say classes[no_of_time_periods]). 我正在使用std :: pair获取每个时间段的开始时间和结束时间,并将所有这些时间段放入数组中(例如,classes [no_of_time_periods])。 Code for it: 代码:
int getClassTimes(int N)
{
int subsets, startTime, endTime;
std::pair<int,int> classes[N];
for(int i=0;i<N;i++)
{
printf("Please Enter Class %d Start Time and End Time:",(i+1));
scanf("%d",&startTime);
scanf("%d",&endTime);
classes[i] = std::make_pair(startTime,endTime);
}
subsets = compute(classes,N);
return subsets;
}
I know I can check if two time periods overlap or not using following condition: 我知道我可以使用以下条件检查两个时间段是否重叠:
if(!(classes[i].first<classes[j].second && classes[i].second>classes[j].first))
But I want to check for more than two time periods. 但是我想检查两个以上的时间段。 Example: 例:
Input:
No_Of_Time_Periods = 5
Time_Period 1 = 1 to 3
Time_Period 2 = 3 to 5
Time_Period 3 = 5 to 7
Time_Period 4 = 2 to 4
Time_Period 5 = 4 to 6
Calculation(Number of subsets of non-overlapping Time Periods):
**Note: If end time of one class is start time of other, they are non-overlapping.
Ex((1 to 3) and (3 to 5) are non-overlapping.)**
(1 to 3)(3 to 5)
(1 to 3)(3 to 5)(5 to 7)
(1 to 3)(4 to 6)
(1 to 3)(5 to 7)
(2 to 4)(4 to 6)
(2 to 4)(5 to 7)
(3 to 5)(5 to 7)
Output:
Total Number of subsets of non-overlapping classes:7
I found the logic and I wrote the code.But while submitting in Online Judge(SPOJ), it says "Time Limit Exceeded".So, Obviously my code was not well optimized. 我找到了逻辑,然后编写了代码。但是在Online Judge(SPOJ)中提交时,它显示“ Time Limited Exceeded”。因此,显然我的代码没有得到很好的优化。 How to achieve this using c++ with better performance? 如何使用具有更好性能的c ++实现此目标? Any Help would be appreciated. 任何帮助,将不胜感激。 Thanks in advance! 提前致谢!
This is so confusing! 真是令人困惑! The overlap test should be like so: 重叠测试应如下所示:
if ((classes[i].second < classes[j].first) || (classes[i].first > classes[j].second))
printf("no overlap");
if (!((classes[i].second < classes[j].first) || (classes[i].first > classes[j].second)))
printf("overlap");
assuming classes[n].first
is always less than classes[n].second
假设classes[n].first
classes[n].second
总是小于classes[n].second
check out interval trees. 检出间隔树。 You will get logarithmic complexity: http://en.wikipedia.org/wiki/Interval_tree 您将获得对数复杂度: http : //en.wikipedia.org/wiki/Interval_tree
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