Length of the longest non-ascending sub-sequence

The longest non-ascending sub-sequence problem is a classic DP problem. A complete description of the problem is given below:

Give you a bunch of sequence a1,a2,a3,a4,a5 ... An. Let you find out one of its oldest firstborn sequences s1,s2,s3,s4 ... Sm. Make S1<=S2<=S3<=S4.........<=SM.

From the problem description, we can change the <= into the longest non-descending sub-sequence by exchanging various other relationship symbols, all of which are just the same as the relationship description. Now, let's just summarize the longest non-ascending subsequence problem to apply to other types.

Since this is a classic DP problem, it is clearly necessary to find out his state transfer equation. First, we use dp[i] to represent the length of the longest non-ascending subsequence of 1--i. Then Dp[i]=max (Dp[j]) +1,j∈[1, I-1]. Apparently dp[1]=1. So go on, oh, just start with the second entry.

The algorithm is described as follows:

1.DP[1] = 1;

2. This iterates through the entire sequence, finding the oldest sequence length from the first item to the current item at a time.

3. Finding the largest of the DP arrays is the oldest sequence length of the entire sequence.

The algorithm complexity is: O (n^2)

1 intNum[n];2 intDp[n];3 4 intLis (intN)5 {6Memset (DP,0,sizeofDP);7     intans;8dp[0] =1;9      for(intI=1; i<n; i++)Ten     { OneAns =Dp[i]; A          for(intj=0; j<i; J + +) -             if(Num[i] <= num[j] && dp[j]>ans) -Ans =Dp[j]; theDp[i] = ans+1; -     } -Ans =0; -      for(intI=0; i<n; i++) +         if(Dp[i] >ans) -Ans =Dp[i]; +     returnans; A}

Length of the longest non-ascending sub-sequence