Special Subsequence
Time Limit: 5000ms
Memory Limit: 32768KB
This problem will be judged on ZJU. Original ID: 334964-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
Author
CHEN, Zhangyi
解题：动态规划+线段树优化
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100100; 4 int tree[maxn<<2],Lisan[maxn],a[maxn],tot,n,d; 5 int query(int L,int R,int lt,int rt,int v){ 6 if(lt <= L && rt >= R) return tree[v]; 7 int mid = (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)); 10 return ret; 11 } 12 void update(int L,int R,int pos,int val,int v){ 13 if(L == R){ 14 tree[v] = max(tree[v],val); 15 return; 16 } 17 int mid = (L + R)>>1; 18 if(pos <= mid) update(L,mid,pos,val,v<<1); 19 if(pos > mid) update(mid + 1,R,pos,val,v<<1|1); 20 tree[v] = max(tree[v<<1],tree[v<<1|1]); 21 } 22 int main(){ 23 while(~scanf("%d%d",&n,&d)){ 24 memset(tree,0,sizeof tree); 25 for(int i = 0; i < n; ++i){ 26 scanf("%d",a + i); 27 Lisan[i] = a[i]; 28 } 29 sort(Lisan,Lisan + n); 30 tot = unique(Lisan,Lisan + n) - Lisan; 31 int ret = 0; 32 for(int i = 0; i < n; ++i){ 33 int L = max(0,(int)(lower_bound(Lisan,Lisan + tot,a[i] - d) - Lisan + 1)); 34 int R = min(tot,(int)(upper_bound(Lisan,Lisan + tot,a[i] + d) - Lisan)); 35 int tmp = 1,pos = lower_bound(Lisan,Lisan + tot,a[i]) - Lisan + 1; 36 if(L <= R) tmp = query(0,tot,L,R,1) + 1; 37 ret = max(ret,tmp); 38 update(0,tot,pos,tmp,1); 39 } 40 printf("%d ",ret); 41 } 42 return 0; 43 }