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

Posted 2021-02-04 zhangbuang

 

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

输入格式：

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.

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.

输入输出样例

输入样例#1： 复制

7 3
10
5
1
12
9
5
14

输出样例#1： 复制

5

---

---

---

 

分析题意

题目大意就是要在排好序的数列中找出两段长度和最大的不重合的区间，并使两个区间中的最大值与最小值的差不大于k。

 

有几种可能的情况，

1.两个区间相邻，

2. 两个区间不相邻

我开始是找的次大和最大的区间然后加起来，对于第2种情况是对的，但对第1种不对，最大区间拆出来一点再向右扩展可能更优

所以有点dp的意思，枚举每个点的前面的点结尾放的最大值的加这个点以后的点开始放的最大值，求max

 

 

code：

 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
",ans);
28 }

 

---

---

---