ZOJ 3349 Special subsequence

Special Subsequencetime limit:5000msmemory limit:32768kbthis problem would be judged onZju. Original id:3349
64-bit integer IO format: %lld Java class name: Main

There a sequence S with n integers, and a are a special subsequence that satisfies | ai-ai-1| <= D (0 <i<=| a|))

Now your task was to find the longest special subsequence of a certain sequence S

Input

There is no more than cases, process till the End-of-file

The first line of all case contains, n and D (1<=n<=100000, 0<=d<=100000000) as in the description.

The second line contains exact n integers, which consist the sequnece S . Each integer was in the range [0,100000000]. There is the blank between each integer.

There is a blank line between the cases

Output

For each case, print the maximum length of special subsequence you can get.

Sample Input

5 21 4 3 6 55 01 2 3 4 5

Sample Output

31

SourceZOJ Monthly, June 2010AuthorCHEN, Zhangyi Problem solving: Dynamic programming + segment Tree optimization

1#include <bits/stdc++.h>2 using namespacestd;3 Const intMAXN =100100;4 inttree[maxn<<2],lisan[maxn],a[maxn],tot,n,d;5 intQueryintLintRintLtintRtintv) {6     if(LT <= L && RT >= R)returnTree[v];7     intMid = (L + R) >>1, ret =0;8     if(LT <= mid) ret = query (l,mid,lt,rt,v<<1);9     if(Rt > Mid) ret = max (Ret,query (Mid +1,r,lt,rt,v<<1|1));Ten     returnret; One } A voidUpdateintLintRintPosintValintv) { -     if(L = =R) { -TREE[V] =Max (tree[v],val); the         return; -     } -     intMid = (L + R) >>1; -     if(POS <= mid) Update (l,mid,pos,val,v<<1); +     if(Pos > Mid) Update (Mid +1,r,pos,val,v<<1|1); -TREE[V] = max (tree[v<<1],tree[v<<1|1]); + } A intMain () { at      while(~SCANF ("%d%d",&n,&d)) { -memset (Tree,0,sizeoftree); -          for(inti =0; I < n; ++i) { -scanf"%d", A +i); -Lisan[i] =A[i]; -         } inSort (Lisan,lisan +n); -tot = unique (Lisan,lisan + N)-Lisan; to         intRET =0; +          for(inti =0; I < n; ++i) { -             intL = Max (0,(int) (Lower_bound (Lisan,lisan + tot,a[i]-D)-Lisan +1)); the             intR = min (Tot, (int) (Upper_bound (Lisan,lisan + tot,a[i] + D)-Lisan)); *             intTMP =1, pos = lower_bound (Lisan,lisan + tot,a[i])-Lisan +1; $             if(L <= R) tmp = query (0, Tot,l,r,1) +1;Panax NotoginsengRET =Max (ret,tmp); -Update0, Tot,pos,tmp,1); the         } +printf"%d\n", ret); A     } the     return 0; +}

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ZOJ 3349 Special subsequence