ZOJ 3349 Special Subsequence

Special Subsequence

Time Limit: 5000ms

Memory Limit: 32768KB

This problem will be judged on ZJU. Original ID: 3349
64-bit integer IO format: %lld      Java class name: Main

There a sequence S with n integers , and A is a special subsequence that satisfies |Ai-Ai-1| <= d ( 0 <i<=|A|))

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

Input

There are no more than 15 cases , process till the end-of-file

The first line of each case contains two integer 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 is in the range [0,100000000] .There is blank between each integer.

There is a blank line between two cases

Output

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

Sample Input

5 2
1 4 3 6 5

5 0
1 2 3 4 5

Sample Output

3
1

 

Source

ZOJ Monthly, June 2010

Author

CHEN, Zhangyi

 

解题：动态规划+线段树优化

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 int tree[maxn<<],Lisan[maxn],a[maxn],tot,n,d;
 int query(int L,int R,int lt,int rt,int v){
     if(lt <= L && rt >= R) return tree[v];
     int mid = (L + R)>>,ret = ;
     if(lt <= mid) ret = query(L,mid,lt,rt,v<<);
     if(rt > mid) ret = max(ret,query(mid + ,R,lt,rt,v<<|));
     return ret;
 }
 void update(int L,int R,int pos,int val,int v){
     if(L == R){
         tree[v] = max(tree[v],val);
         return;
     }
     int mid = (L + R)>>;
     if(pos <= mid) update(L,mid,pos,val,v<<);
     if(pos > mid) update(mid + ,R,pos,val,v<<|);
     tree[v] = max(tree[v<<],tree[v<<|]);
 }
 int main(){
     while(~scanf("%d%d",&n,&d)){
         memset(tree,,sizeof tree);
         for(int i = ; i < n; ++i){
             scanf("%d",a + i);
             Lisan[i] = a[i];
         }
         sort(Lisan,Lisan + n);
         tot = unique(Lisan,Lisan + n) - Lisan;
         int ret = ;
         for(int i = ; i < n; ++i){
             int L = max(,(int)(lower_bound(Lisan,Lisan + tot,a[i] - d) - Lisan + ));
             int R = min(tot,(int)(upper_bound(Lisan,Lisan + tot,a[i] + d) - Lisan));
             int tmp = ,pos = lower_bound(Lisan,Lisan + tot,a[i]) - Lisan + ;
             if(L <= R) tmp = query(,tot,L,R,) + ;
             ret = max(ret,tmp);
             update(,tot,pos,tmp,);
         }
         printf("%d\n",ret);
     }
     return ;
 }