P3143 [USACO16OPEN]Diamond Collector S

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本文介绍了一个关于如何在两个陈列架上最大化展示相似大小钻石的问题。Bessie希望钻石之间的大小差值不超过K，通过排序和计算每个位置的最大显示数，可以找到最佳展示方案。算法包括对钻石进行排序，计算每个位置左侧和右侧的最大距离，然后枚举中间值求解最大总和。

Bessie the cow, always a fan of shiny objects, has taken up a hobby of mining diamonds in her spare time! She has collected NN diamonds (N \leq 50,000N≤50,000) of varying sizes, and she wants to arrange some of them in a pair of display cases in the barn.

Since Bessie wants the diamonds in each of the two cases to 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 KK (two diamonds can be displayed together in the same case if their sizes differ by exactly KK). Given KK, please help Bessie determine the maximum number of diamonds she can display in both cases together.

奶牛Bessie很喜欢闪亮亮的东西（Baling_Baling），所以她喜欢在她的空余时间开采钻石！她现在已经收集了N颗不同大小的钻石（N<=50,000），现在她想在谷仓的两个陈列架上摆放一些钻石。

Bessie想让这些陈列架上的钻石保持相似的大小，所以她不会把两个大小相差K以上的钻石同时放在一个陈列架上（如果两颗钻石的大小差值为K，那么它们可以同时放在一个陈列架上）。现在给出K，请你帮Bessie确定她最多一共可以放多少颗钻石在这两个陈列架上。

输入格式
The first line of the input file contains NN and KK (0 \leq K \leq 1,000,000,0000≤K≤1,000,000,000).

The next NN lines each contain an integer giving the size of one of the

diamonds. All sizes will be positive and will not exceed 1,000,000,0001,000,000,000.

输出格式
Output a single positive integer, telling the maximum number of diamonds that

Bessie can showcase in total in both the cases.

算是思维题吧
先sort一趟，然后求一下每个点到左右最远的一个长度
对于每个点取左边的最大值，右边同理
枚举中间值，求左侧和右侧的最大加和

#include<iostream>
#include<algorithm>
using namespace std;
int a[50010];
int z[50010];
int y[50010];
int main()
{
    int n,k,l=1,ma=0;
    cin>>n>>k;
    for(int i=1;i<=n;i++) cin>>a[i];
    sort(a+1,a+n+1);
    for(int i=1;i<=n;i++)
    {
        while(a[i]-a[l]>k) l++;
        z[i]=i-l+1;
    }
    for(int i=n,l=n;i;i--)
    {
        while(a[l]-a[i]>k) l--;
        y[i]=l-i+1;
    }
    for(int i=1;i<=n;i++) z[i]=max(z[i],z[i-1]);
    for(int i=n;i>=1;i--) y[i]=max(y[i],y[i+1]);
    for(int i=1;i<=n;i++) ma=max(ma,z[i]+y[i+1]);
    cout<<ma<<endl;
}