Question

I want to know the best way to implement this (approach). I will bee given set of coordinates (x,y). I will then be queried based on those coordinates like

1.C a b => where a and b are integer indexes in the initial set of coordinates. So I need to output the number of points that are in 1st ,2nd ,3rd and fourth quadrant that are in index range a to b.

2.X a b => where a and b are integer indexes in the initial set of coordinates. So I need to mirror the ath to bth indexed coordinates along x axis.

3.Y a b => where a and b are integer indexes in the initial set of coordinates. So I need to mirror the ath to bth indexed coordinates along y axis.

there can be at most 100000 coordinates or points and 500000 such queries on them.

I tried a brute force method looping through on every query but it had TLE(Time Limit Exceeded).

What should I do in such type of questions?

Here is my code

#include <iostream>
#include <stdio.h>

using namespace std;

char flipX[4] = { 3, 2, 1, 0 };
char flipY[4] = { 1, 0, 3, 2 };

int main(int argc, char **argv)
{
    int n,x,y;
    char c[100000];
    scanf("%d",&n);
    //coord *c=new coord[n];
    for(int i=0;i<n;i++)
    {
        scanf("%d",&x);
        scanf("%d",&y);
        if(x<0 && y<0)
            c[i]=2;
        else if(x>0 && y>0)
            c[i]=0;
        else if(x>0 && y<0)
            c[i]=3;
        else
            c[i]=1;
    }

    int q,i,j,a,cnt[4];
    char ch;
    scanf("%d",&q);
    while(q--)
    {
        //cout<<"q:"<<q<<endl;
        cin>>ch;
        scanf("%d",&i);
        scanf("%d",&j);
        i--;j--;
        if(ch=='X')
        {
            //case 'X':
                for(a=i;a<=j;a++)
                    c[a]=flipX[c[a]];
            //  break;
        }
        else if(ch=='Y')
        {
            //case 'Y':
                for(a=i;a<=j;a++)
                    c[a]=flipY[c[a]];
            //  break;
        }   
        else if(ch=='C')
        {
                cnt[0]=cnt[1]=cnt[2]=cnt[3]=0;
                for(a=i;a<=j;a++)
                {
                    cnt[c[a]]++;
                }
                printf("%d %d %d %d\n",cnt[0],cnt[1],cnt[2],cnt[3]);
        }
    }
    return 0;
}

Please Help.

Answer 1

Ok. Lazy segment trees did the trick. Thanx everyone for your help