bzoj4582 [Usaco2016 Open]Diamond Collector

bzoj4582 [Usaco2016 Open]Diamond Collector

​​ 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#include#include#define ll long longusing 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;}

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