Nlogn solution of longest ascending subsequence sequence in DP practice

Openjudge 2757: Longest ascending sub-sequence

Total time limit:

2000ms

Memory Limit:

65536kB

Describe

A number of sequence bi, when B1 < B2 < ... < BS , we call this sequence ascending.  For a given sequence ( A1, A2, ..., an), We can get some ascending sub-sequences ( ai1, ai2, ..., AiK), here 1 <= i1 < i2 < ... < IK <= N. For example, for sequences (1, 7, 3, 5, 9, 4, 8), there are some ascending sub-sequences of it, such as (1, 7), (3, 4, 8) and so on. The longest length of these subsequence sequences is 4, such as subsequence (1, 3, 5, 8).

Your task is to find the length of the longest ascending sub-sequence for a given sequence.

Input

The first line of input is the length of the sequence n (1 <= n <= 1000). The second line gives the n integers in the sequence, with values ranging from 0 to 10000.

Output

The length of the longest ascending sub-sequence.

Sample input

71 7 3 5 9 4 8

Sample output

4

1 /*Do this problem again because another topic is to reverse the use of this C array, here to review*/2#include <iostream>3 using namespacestd;4#include <cstdio>5#include <cstring>6 intN;7 #defineN 10108 intc[n],f[n],ans=-N,a[n];9 intSearchintLintRintk)Ten { One     if(L==R)returnl; A     intMid= (l+r+1) >>1;/*and +1 here.*/ -     if(c[mid]>=k)returnSearch (l,mid-1, k);/*the mid-1 here is guaranteed to be ascending sequence*/ -     Else returnsearch (mid,r,k); the } - intMain () - { -scanf"%d",&n); +      for(intI=1; i<=n;++i) -scanf"%d",&a[i]); +Memset (F,0,sizeof(f)); AMemset (c,127,sizeof(c)); at      for(intI=1; i<=n;++i) -     { -F[i]=search (0, I,a[i]) +1;/*the specific two-part process here is best to do a manual simulation to prevent errors*/ -c[f[i]]=min (A[i],c[f[i]]); -ans=Max (F[i],ans); -     } inprintf"%d\n", ans); -     return 0; to}

Nlogn solution of longest ascending subsequence sequence in DP practice