codeforces985E 2000 points dp

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Meaning of the questions:

n balls, each ball there are a number of AI, two integer k and d. Now you want to stack divided into a plurality of n balls, there is no limit to the number of stacks, each stack comprising at least k balls, each stack number does not exceed the maximum value minus minimum d. You can ask whether such stock piles.

Data range: 1 <= k <= n <= 5e5, 0 <= d <= 1e9, 1 <= ai <= 1e9.

answer:

When think the current number of pile points more likely to make the results better, it should be aware of dp.

dp [i] = 1 denotes the front i is the number of points can be made to meet the requirements of the stack, dp [i] = 0 indicates the number i is not carried out prior to meet the requirements of the sub-stack.

0 is a valid number, which is the initial state, i.e. dp [0] = 1.

This number n sort, from small to large to traverse.

Two legal limits can be drawn interval [L, R], L satisfies aL> = ai - d smallest subscript, R <= i - k.

If dp [j] = 1 and j∈ [L, R], the dp [i] = 1.

Get dp [J] is the process in the course of seeking the array dp [L. R] interval maximum value.

I did not open dp array, because I keep the state tree line inside.

Experience:

When setting the initial state forget to change a parameter, resulting in 2A.

Not enough stable.

This is a problem when you realize dp would have done, different levels of players aware of the time, I was not one, had to think for a while.

Indeed raise the level of some of the previously encountered this problem, see a solution to a problem you think about it, and now can do independent of (although looked wrong test cases).

Code:

#include<bits/stdc++.h>
using namespace std ;
typedef long long ll ;
const int maxn = 5e5 + 5 ;
int n , k , d ;
int a[maxn] ;
bool max1[maxn << 2] ;
int ls(int x)
{
    return x << 1 ;
}
int rs(int x)
{
    return x << 1 | 1 ;
}
void update(int id , int l , int r , int x , bool y)
{
	int mid = (l + r) / 2 ;
	if(l == r && l == x)
	{
		max1[id] = y ;
		return ;
	}
	if(x <= mid)
	  update(ls(id) , l , mid , x , y) ;
	else
	  update(rs(id) , mid + 1 , r , x , y) ;
	max1[id] = max(max1[ls(id)] , max1[rs(id)]) ;
}
bool query(int id , int l , int r , int x , int y)
{
    if(x > y)  return 0 ;
	bool ans = 0 ;
	int mid = (l + r) / 2 ;
	if(x <= l && r <= y)
	  return max1[id] ;
	if(x <= mid) ans = max(ans , query(ls(id) , l , mid , x , y)) ;
	if(y > mid)  ans = max(ans , query(rs(id) , mid + 1 , r , x , y)) ;
	return ans ;
}
void solve(int i)
{
	int l = lower_bound(a + 1 , a + i + 1 , a[i] - d) - a ;
	int r = i - k ;
	l -- ;
	l = max(l , 0) ;
	bool x = query(1 , 0 , n , l , r) ;
	if(x)  update(1 , 0 , n , i , 1) ;
}
int main()
{
	scanf("%d%d%d" , &n , &k , &d) ;
	for(int i = 1 ; i <= n ; i ++)  scanf("%d" , &a[i]) ;
	sort(a + 1 , a + n + 1) ;
	update(1 , 0 , n , 0 , 1) ;
	for(int i = 1 ; i <= n ; i ++)  
	  solve(i) ;
	if(query(1 , 0 , n , n , n))  printf("YES\n") ;
	else  printf("NO\n") ;
	return 0 ;
}