BZOJ 4582: [Usaco2016 Open]Diamond Collector

4582: [Usaco2016 Open]Diamond Collector

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 164  Solved: 111

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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 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 K (two diamonds can be displayed together in the same case if
their 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 contain an integer giving the size of one
of the diamonds. All sizes will be positive and will not exceed 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

并不知道应该分到哪一类

感觉上get了一种思想

对于不要求连续的子序列问题

可以改变元素的位置，从而大幅度简化复杂度（不要求顺序）

对于连续子序列

可以采用分治思想考虑把序列拆解

可以考虑序列左右端点对序列的影响

以上内容纯属博主唯心主义（当然博主不是唯心主义者）的归纳

这道题可以用以上思想解决

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<complex>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f*x;
}
int n,m,i,r,a[50005],f[50005],g[50005],ans;
int main()
{
n=read();m=read();
for(i=1;i<=n;i++)a[i]=read();
sort(a+1,a+n+1);r=1;
for(i=1;i<=n;i++)
{
while(a[r+1]-a[i]<=m&&r<n)r++;
f[i]=r-i+1;
}
for(i=n;i>=1;i--)g[i]=max(g[i+1],f[i]);
ans=0;
for(i=1;i<=n;i++)ans=max(ans,f[i]+g[i+f[i]]);
printf("%d",ans);
return 0;
}