I . K-gap Subsequence

题目大意

A subsequence of a given sequence of integers is a subset of the values in the sequence in the same order. A kk-gap subsequence of a sequence of integers is a subsequence such that consecutive elements in the subsequence differ by at least kk. For example, the sequence

3, 12, 8, 4, 2, 5, 1, 9, 8, 6, 5, 1, 7

has a 4-gap subsequence of 3, 12, 8, 4, 9, 5, 1 and 7 (highlighted in bold above) since |3 - 12|, |12 - 8|, |8 - 4|, |4 - 9|, |9 - 5|, |5 - 1| and |1 - 7| are all 4 or greater.

Given a sequence and a value of kk, determine the length of the longest kk-gap subsequence.

Input

The first input line contains two space separated integers: nn (1≤n≤300,0001≤n≤300,000), indicating the length of the sequence, and kk (1≤k≤1091≤k≤109), indicating the value of kk for the input case.

The following input line contains nn space separated integers. The iith of these integers is aiai (1≤ai≤1091≤ai≤109 ), the ith value of the input sequence.

Output

Print the length of the longest kk-gap subsequence of the input sequence.

Samples

Input 复制

10 2
1 2 3 2 1 3 1 3 5 6

Output

7

Input 复制

5 12
3 7 14 20 32

Output

3

Input 复制

13 4
3 12 8 4 2 5 1 9 8 6 5 1 7

Output8

题解

这个题意就是给你n个数，然后让你从中选出一些子序列，使得相邻的两个数的绝对值小于k,这个题是一个线段树+dp，首先我们把这些数给他排序+离散化一下，然后我们操作了一个数，那么下一个数就是
从这些符合条件的已经操作晚的数中选一个
Code

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=5e6+100;
struct node{
    int l,r;
    int ma;
}t[maxn];
int a[maxn],b[maxn];
void push(int p){
    t[p].ma=max(t[2*p].ma,t[2*p+1].ma);
}
void build(int p,int l,int r){
    t[p].l=l;
    t[p].r=r;
    t[p].ma=0;
    if(l==r){
        return ;
    }
    int mid=(l+r)/2;
    build(2*p,l,mid);
    build(2*p+1,mid+1,r);
    push(p);
} 
void update(int p,int l,int x){
    if(t[p].l==t[p].r){
        t[p].ma=x;
        return ;
    }
    int mid=(t[p].l+t[p].r)/2;
    if(l<=mid){
        update(2*p,l,x);
    }
    else{
        update(2*p+1,l,x); 
    }
    push(p);
}
int query(int p,int l,int r){
    if(t[p].l>=l&&t[p].r<=r){
        return t[p].ma;
    }
    int mid=(t[p].l+t[p].r)/2;
    int ans=0;
    if(l<=mid){
        ans=max(ans,query(2*p,l,r));
    }
    if(r>mid){
        ans=max(ans,query(2*p+1,l,r));
    }
    return ans;
}
int main(){
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>a[i]; 
        b[i]=a[i];
    }
    sort(b+1,b+n+1);
    int nn=unique(b+1,b+n+1)-(b+1);
    build(1,1,nn);
    int ans=0;
    for(int i=1;i<=n;i++){
        int l=upper_bound(b+1,b+nn+1,a[i]-k)-b-1;
        int r=lower_bound(b+1,b+nn+1,a[i]+k)-b;
        int cnt=0;
        if(l>=1){
            cnt=max(cnt,query(1,1,l));
        }
        if(nn>=r){
            cnt=max(cnt,query(1,r,nn));
        }
        ans=max(ans,++cnt);
        //cout<<ans<<" cnt="<<cnt<<" "<<l<<" "<<r<<endl;
        int t=lower_bound(b+1,b+nn+1,a[i])-b;
        update(1,t,cnt);
    }
    cout<<ans<<endl;
}