TYVJ 1049 Longest non-descending sub-sequence N^2/nlogn

P1049Maximum non-descending subsequence time: 1000ms/Space: 131072kib/java Class Name: main describes the length of the longest non-descending subsequence input format the first behavior N, which represents the number of n
Second row n number of output format longest non-descending subsequence length test sample 1 input

3
1 2 3

Output

3

Note N is less than 5000
For each num <=maxint Test instructions: Chinese test instructions: No descent is >= n^n dp[i] Indicates the length of the longest non-descending subsequence of the number of previous I

1 /******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^    ^ ^5 o o6 ******************************/7 //#include <bits/stdc++.h>8#include <iostream>9#include <cstring>Ten#include <cstdio> One#include <map> A#include <algorithm> -#include <cmath> - #definell Long Long the #definePI ACOs (-1.0) - #defineMoD 1000000007 - using namespacestd; - intN; + inta[5005]; - intdp[5005]; + intMain () A { at      while(~SCANF ("%d",&N)) { -Memset (DP,0,sizeof(DP)); -      for(intI=1; i<=n;i++) -         { -scanf"%d",&a[i]); -dp[i]=1; in         } -     intmaxn=1; to     intans=0; +      for(intI=1; i<=n;i++) -     { themaxn=1; *          for(intj=1; j<i;j++) $         {Panax Notoginseng             if(a[j]<=a[i]&&dp[j]+1>MAXN) -maxn=dp[j]+1; the         } +dp[i]=MAXN; Aans=Max (ANS,MAXN); the     } +cout<<ans; -     } $     return 0; $}

Nlogn using Upper_bound and the longest ascending subsequence to judge the boundary of different attention

Ans[i] represents the longest non-descending value of the last digit of length I

1 /******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^    ^ ^5 o o6 ******************************/7 //#include <bits/stdc++.h>8#include <iostream>9#include <cstring>Ten#include <cstdio> One#include <map> A#include <algorithm> -#include <cmath> - #definell Long Long the #definePI ACOs (-1.0) - #defineMoD 1000000007 - using namespacestd; - intN; + inta[5005]; - intMain () + { A      while(~SCANF ("%d",&N)) { atMemset (DP,0,sizeof(DP)); -      for(intI=0; i<n;i++) -scanf"%d",&a[i]); -     intmaxn=1; -     intans[5005]; -     inttop=1; inans[1]=a[0]; -      for(intI=1; i<n;i++) to     { +        if(a[i]>=Ans[top]) -ans[++top]=A[i]; the        Else *        { $            intPos=upper_bound (Ans,ans+top,a[i])-ans;//points to the position of the first element greater than A[i]Panax Notoginsengans[pos]=a[i];//Update -        } the     } +cout<<top<<Endl;; A     } the     return 0; +}

TYVJ 1049 Longest non-descending sub-sequence N^2/nlogn