@sanchit nested loops does not mean TLE
I have just traversed the whole process's time array once for each elapse
time calculations ie O(n) and O(n^2) in total. I am not able to figure out
a faster method or a DP approach. Suggestions awaited
On Mon, Mar 21, 2011 at 11:20 PM, sanchit mittal <sm14it...@gmail.com>wrote:
> with so many nested loops u r obvcly goin to get tle....
>
>
> On Mon, Mar 21, 2011 at 11:07 PM, ayush bhutani <ayushbhut...@gmail.com>wrote:
>
>> I have got a better algo
>> But I am also getting TLE
>> #include<stdio.h>
>> int main(){
>> int n,i,j;
>> long t[50000];
>> long elapse[50000];
>> scanf("%d",&n);
>> for(i=0;i<n;i++){
>> scanf("%ld",&t[i]);
>> elapse[i]=0;
>> }
>> for(i=0;i<n;i++){
>>
>> for(j=0;j<n;j++){
>> if(j!=i){
>> if(t[j]<t[i]){
>> elapse[i] +=t[j];
>> }
>> else{
>> if(j<i){
>> elapse[i]+=t[i];
>> }
>> else{
>> elapse[i] +=(t[i]-1);
>> }
>> }
>> }
>> else{
>> elapse[i] +=t[i];
>> }
>> }
>> }
>> for(i=0;i<n;i++){
>> printf("%ld\n",elapse[i]);
>> }
>>
>> return 0;
>> }
>>
>>
>> On 3/21/11, Akshata Sharma <akshatasharm...@gmail.com> wrote:
>> > Which data structure to use for getting it accepted?? This algorithm
>> > gives tle. Using scanf, printf is also resulting in tle.
>> >
>> > On Mar 21, 7:06 pm, radha krishnan <radhakrishnance...@gmail.com>
>> > wrote:
>> >> U have to use Some Advanced Data Structures for this problem
>> >> :P
>> >>
>> >>
>> >>
>> >> On Mon, Mar 21, 2011 at 3:02 PM, saurabh singh <saurab...@gmail.com>
>> >> wrote:
>> >> > using scanf and printf and still tle,I am not pretty sure how malloc
>> or
>> >> > new
>> >> > arrays can speed up execution?
>> >>
>> >> > On Mon, Mar 21, 2011 at 2:25 PM, sanchit mittal <sm14it...@gmail.com
>> >
>> >> > wrote:
>> >>
>> >> >> use scanf n printf instead of cin n cout,
>> >> >> malloc array of structures after reading value of n if working in c
>> >> >> else
>> >> >> use new in cpp
>> >> >> rest i guess...is ok
>> >> >> On Sun, Mar 20, 2011 at 11:53 PM, ankit sambyal
>> >> >> <ankitsamb...@gmail.com>
>> >> >> wrote:
>> >>
>> >> >>> I worked on this problem but cud not get a more efficient algo than
>> >> >>> yours.
>> >> >>> Plz get back 2 me if u find a better algo.
>> >>
>> >> >>> On Sun, Mar 20, 2011 at 3:24 AM, Akshata Sharma
>> >> >>> <akshatasharm...@gmail.com> wrote:
>> >>
>> >> >>>> I tried to solve this problem
>> >> >>>>https://www.spoj.pl/problems/RRSCHED/
>> >>
>> >> >>>> I am getting TLE!! How can I improve my code??
>> >>
>> >> >>>> #include<iostream>
>> >> >>>> #include<stdio.h>
>> >>
>> >> >>>> using namespace std;
>> >>
>> >> >>>> struct process
>> >> >>>> {
>> >> >>>> long time;
>> >> >>>> int finished;
>> >> >>>> long elapsed_time;
>> >> >>>> };
>> >>
>> >> >>>> int main()
>> >> >>>> {
>> >> >>>> long n,sum=0;
>> >> >>>> cin>>n;
>> >> >>>> struct process prss[50000];
>> >> >>>> for(long i=0;i<n;i++)
>> >> >>>> {
>> >> >>>> scanf("%ld",&prss[i].time);
>> >> >>>> prss[i].finished=0;
>> >> >>>> sum+=prss[i].time;
>> >> >>>> }
>> >> >>>> long index=0;
>> >> >>>> for(long k=1;k<=sum;k++)
>> >> >>>> {
>> >> >>>> while(prss[index].finished==1)
>> >> >>>> index++;
>> >>
>> >> >>>> prss[index].time--;
>> >>
>> >> >>>> if(prss[index].time==0)
>> >> >>>> {
>> >> >>>> prss[index].finished=1;
>> >> >>>> prss[index].elapsed_time=k;
>> >> >>>> }
>> >>
>> >> >>>> index++;
>> >> >>>> if(index==n)
>> >> >>>> index=0;
>> >> >>>> }
>> >>
>> >> >>>> for(long i=0;i<n;i++)
>> >> >>>> printf("%ld\n",prss[i].elapsed_time);
>> >> >>>> return 0;
>> >> >>>> }
>> >>
>> >> >>>> --
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>> >> >> --
>> >> >> Sanchit Mittal
>> >> >> Second Year Undergraduate
>> >> >> Computer Engineering
>> >> >> Delhi College of Engineering
>> >> >> ph- +919582414494
>> >>
>> >> >> --
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>> >>
>> >> > --
>> >> > Saurabh Singh
>> >> > B.Tech (Computer Science)
>> >> > MNNIT ALLAHABAD
>> >>
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>
>
> --
> Sanchit Mittal
> Second Year Undergraduate
> Computer Engineering
> Delhi College of Engineering
> ph- +919582414494
>
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