im sorry .. y<k fails :D
On 6/11/11, Radhika Renganathan <radi.coo...@gmail.com> wrote:
> yea.. now got ac.. :) mistake was k==y is also possible but x<k fails.. so
> got WA .. thank u :)
>
> On Sat, Jun 11, 2011 at 2:39 PM, keyan karthi
> <keyankarthi1...@gmail.com>wrote:
>
>> k=query(x,y-1)
>> if(k==x)
>> count++
>> with this change ur code ACs :)
>>
>>
>> On Sat, Jun 11, 2011 at 1:24 PM, Radhika Renganathan <
>> radi.coo...@gmail.com> wrote:
>>
>>> i did the same prob wit range maximum query.. but im repeatedly
>>> getting wrong answer.. im stuck with this prob for a long time.. pls
>>> help..
>>>
>>> my code:
>>>
>>> #include<iostream>
>>> using namespace std;
>>> #include<stdlib.h>
>>> #include<stdio.h>
>>> int A[50010];
>>> int M[9999999];
>>> void initialize(int node, int b, int e)
>>> {
>>> if (b == e)
>>> M[node] = b;
>>> else
>>> {
>>> initialize(2 * node, b, (b + e) / 2);
>>> initialize(2 * node + 1, (b + e) / 2 + 1,e);
>>> if (A[M[2 * node]] >= A[M[2 * node + 1]])
>>> M[node] = M[2 * node];
>>> else
>>> M[node] = M[2 * node + 1];
>>> }
>>> }
>>> int query(int node, int b, int e, int i, int j)
>>> {
>>> int p1, p2;
>>> if (i > e || j < b)
>>> return -9999;
>>> if (b >= i && e <= j)
>>> return M[node];
>>> p1 = query(2 * node, b, (b + e) / 2,i, j);
>>> p2 = query(2 * node + 1, (b + e) / 2 + 1, e, i, j);
>>> if (p1 == -9999)
>>> return p2;
>>> if (p2 == -9999)
>>> return p1;
>>> if (A[p1] >= A[p2])
>>> return p1;
>>> return p2;
>>>
>>> }
>>>
>>> int main()
>>> {
>>> int n,i,t,j,count=0,k,size;
>>>
>>> scanf("%d%d",&n,&t);
>>>
>>> for (i=1;i<=n;i++)
>>> scanf("%d",&A[i]);
>>>
>>> initialize(1,1,n);
>>> for(i=0;i<t;i++)
>>> {
>>> int x,y;
>>> scanf("%d%d",&x,&y);
>>> k=query(1,1,n,x,y);
>>> if(!(x<k && k<y))
>>> count++;
>>> }
>>> printf("%d",count);
>>> return 0;
>>> }
>>>
>>>
>>> On 6/11/11, KK <kunalkapadi...@gmail.com> wrote:
>>> > Search Topcoder Tutorial Range Minimum Query @ Google...
>>> > By few intuitive changes u can implement Range Maximum Query as
>>> > well...
>>> >
>>> > --
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>>> >
>>> >
>>>
>>>
>>> --
>>> .... radhika .. :)
>>>
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>>>
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>
>
>
> --
> .... radhika .. :)
>
--
.... radhika .. :)
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