P3143 [USACO16OPEN]钻石收藏家Diamond Collector（伸缩法）

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 NNN diamonds (N≤50,000N \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 KKK (two diamonds can be displayed together in the same case if their sizes differ by exactly KKK). Given KKK, 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。

输入输出格式

The first line of the input file contains NNN and KKK (0≤K≤1,000,000,0000 \leq K \leq 1,000,000,0000≤K≤1,000,000,000).

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

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

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

Bessie can showcase in total in both the cases.

输入输出样例

7 3
10
5
1
12
9
5
14

5

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 int a[50001],n,k,ans=0;
 6 int l[50001],r[50001];
 7 int main()
 8 {
 9     scanf("%d%d",&n,&k);
10     for (int i=1;i<=n;++i) scanf("%d",&a[i]);
11     sort(a+1,a+n+1);
12     int h=1; l[1]=1;
13     for (int i=2;i<=n;++i)
14     {
15         while (a[i]-a[h]>k) h++;
16         l[i]=max(l[i-1],i-h+1);
17     }
18     r[n]=1;
19     h=n;
20     for (int i=n-1;i>=1;--i) 
21     {
22         while (a[h]-a[i]>k) h--;
23         r[i]=max(r[i+1],h-i+1);
24     }
25     for (int i=1;i<n;++i)
26         ans=max(ans,l[i]+r[i+1]);
27     printf("%d\n",ans);
28 }