zoj 3349 Special Subsequence 【离散化二分 + 线段树优化dp】

Special Subsequence

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Time Limit: 5 Seconds       Memory Limit: 32768 KB

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

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

定义D序列：序列里相邻元素差的绝对值不超过d。

题意：给定n个元素组成的序列a[]，求最长的D序列。

思路：考虑升序排列的序列，那么则有dp[i] = max(dp[j]) (l <= j <= r)。其中a[l]为大于或者等于a[i]-d的第一个数，a[r]为小于或者等于a[i]+d的最后一个数。维护升序序列，并不丢失序列元素的下标，可以先将坐标离散化。考虑对当前元素a[i]的处理，我们只需要处理出a[l] >= a[i] - d，a[r] <= a[i] + d以及a[i]在离散化序列中的下标L, R, P，用线段树维护[L, R]区间的最大值，更新到P对应的区间上。只要在这个过程中维护序列长度即可。

偷懒用了C++lower_boundWA了两次。还是自己手写好。。。

AC代码：

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