bzoj4582 [Usaco2016 Open]Diamond Collector

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这是一篇关于计算问题的博客，Bessie奶牛收集了不同大小的钻石，她想将它们分组展示，每组内的钻石大小差异不超过K。任务是找出能展示的最大钻石数量。给定钻石数量N、差异限制K和每个钻石的大小，需要找到两个互不相交的子序列，使子序列差值不超过K，最大化子序列长度之和。输入包含N、K和钻石大小，输出是最大展示数量。

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http://www.elijahqi.win/archives/3050
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 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

Input

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
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
HINT

Source

Silver鸣谢frank_c1提供翻译

先排序

从右向左每个点维护向右最远能延伸的距离然后因为是两个然后我再维护下前缀最远延伸所以直接把两个相加每次取最大即可

#include<cstdio>
#include<cctype>
#include<algorithm>
#define ll long long
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
const int N=50050;
int dp[N];int n,k,d[N];
int main(){
    freopen("bzoj4582.in","r",stdin);
    n=read();k=read();for (int i=1;i<=n;++i) d[i]=read();
    sort(d+1,d+n+1);int nowr=n,maxr=0,now=1,ans=0;
    for (int i=n-1;i;--i){
        while(d[nowr]-d[i]>k) --now,maxr=max(maxr,dp[nowr]),--nowr;
        ++now;dp[i]=now;ans=max(ans,now+maxr);
    }printf("%lld\n",ans);
    return 0;
}