Question

我使用std :: pair来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在一个数组中（比如类[no_of_time_periods]）。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;
}

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))

但我想检查两个以上的时间段。例如：

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

我找到了逻辑并且编写了代码。但是在在线评判（SPOJ）中提交时，它说“超出时间限制”。所以，显然我的代码没有得到很好的优化。如何使用c ++实现更好的性能？任何帮助，将不胜感激。提前谢谢！

Answer 1

这太令人困惑了！重叠测试应该是这样的：

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");

假设classes[n].first始终小于classes[n].second

Answer 2

检查间隔树。您将获得对数的复杂性：http://en.wikipedia.org/wiki/Interval_tree

如何检查c ++中两个或多个时间段是否相互重叠？

2 个答案: