模拟 [bzoj 4582] Diamond Collector

这道是权限，所以我粘个题面

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 198 Solved: 134
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Description
Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare
time! She has collected N diamonds (N≤50,000) of varying sizes, and she wants to arrange some of th
em in a pair of display cases in the barn.Since Bessie wants the diamonds in each of the two cases t
o be relatively similar in size, she decides that she will not include two diamonds in the same case
if their sizes differ by more than K (two diamonds can be displayed together in the same case if th
eir sizes differ by exactly K). Given K, please help Bessie determine the maximum number of diamonds
she can display in both cases together.
给定长度为N的数列a，要求选出两个互不相交的子序列（可以不连续），满足同一个子序列中任意两个元素差的绝
对值不超过K。最大化两个子序列长度的和并输出这个值。1 ≤ N ≤ 50000, 1 ≤ a_i ≤ 10 ^ 9, 0 ≤ K ≤ 10^ 9
Input
The first line of the input file contains N and K (0≤K≤1,000,000,000). The next NN lines each cont
ain an integer giving the size of one of the diamonds. All sizes will be positive and will not excee
d 1,000,000,000
Output
Output a single positive integer, telling the maximum number of diamonds that Bessie can showcase in
total in both the cases.
Sample Input
7 3
10
5
1
12
9
5
14
Sample Output
5
HINT

我们可以搞出对于每个点，他左边的最大长度，他右边的最大长度。
只要枚举左端点，找出最右面的端点，不断更新，反正乱搞出向左向右的最大值，之后重新枚举每个区间，找他左或右的最大值，更新答案。
真没自己想出来，有些貌似数据结构的题，其实会很简单。。

#pragma GCC optimize("O3")
#include
#include
#include
#include
#include
#define N 50005
using namespace std;
int n,k,a[N],L[N],R[N];
int main()
{   
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    if(a[n]-a[1]<=2*k){cout<return 0;}
    for(int l=1,r=1;l<=n;l++)
    {
        while(r<=n&&a[r]-a[l]<=k){R[l]=max(R[l],r-l+1);r++;}
        r--;R[l]=max(R[l],r-l+1);L[r]=r-l+1;
    }
    for(int i=1;i<=n;i++)if(L[i-1]>L[i])L[i]=L[i-1];
    for(int i=n;i>=1;i--)if(R[i+1]>R[i])R[i]=R[i+1];
    int ans=0;
    for(int l=1,r=1;l<=n;l++)
    {
        while(r<=n&&a[r]-a[l]<=k)r++;
        r--;ans=max(ans,r-l+1+max(L[l-1],R[r+1]));
    }
    cout<