Discussion @ 1203. Scientific Conference @ Timus Online Judge

Sanatbek_Matlatipov It is easy, just sort and one linear cycle [1] // Problem 1203. Scientific Conference 12 Sep 2015 12:15

1. Sort it by endTime, if endTime equals another endTime then sort them by startTime.
2. Open one iteration from 2 to n, init just check that, is endTime<starTime, if so then ans++;
3. print ans; endTime and startTimes is one array with unchanged same index...

Sorry for poor English

Dhruv Somani Re: It is easy, just sort and one linear cycle // Problem 1203. Scientific Conference 1 May 2016 22:22

I don't think sorting by startTime is necessary. I think we should reverse-sort by startTime.

Discussion of Problem 1203. Scientific Conference