[Usaco2016 Open]Diamond Collector

题目描述

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

输入格式

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

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.

样例输入

7 3
10
5
1
12
9
5
14

样例输出

5

提示

Silver鸣谢frank_c1提供翻译

思路

我或许被洛谷骗了，这不是splay的题。。。而是某种神奇的贪心。

代码实现

 #include<cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 int s[],n,k,ans=;
 int l[],r[];
 int main(){
     scanf("%d%d",&n,&k);
     for (int i=;i<=n;++i) scanf("%d",&s[i]);
     sort(s+,s+n+);
     int h=; l[]=;
     for (int i=;i<=n;++i){
         while (s[i]-s[h]>k) h++;
         l[i]=max(l[i-],i-h+);
     }
     r[n]=,h=n;
     for (int i=n-;i>=;--i){
         while(s[h]-s[i]>k) h--;
         r[i]=max(r[i+],h-i+);
     }
     for (int i=;i<n;++i) ans=max(ans,l[i]+r[i+]);
     printf("%d\n",ans);
 }