"Dynamic planning + binary search" poj2533&poj1631 longest ascending subsequence (LIS)

POJ2533 bare lis, Time complexity O (n^2)

1#include <iostream>2#include <cstdio>3 using namespacestd;4 Const intmaxn= ++5;5 intA[MAXN];6 intDP[MAXN];7 intN,ans;8 9 intMain ()Ten { Onescanf"%d",&n); A      for(intI=0; i<n;i++) -     { -scanf"%d",&a[i]); thedp[i]=1; -     } -ans=-1; -      for(intI=1; i<n;i++) +     { -          for(intj=0; j<i;j++) +             if(A[j]<a[i] && dp[j]+1>Dp[i]) A             { atdp[i]=dp[j]+1; -             } -         if(dp[i]>ans) -             { -ans=Dp[i]; -             } in     } -cout<<ans<<Endl; to     return 0; +}

POJ1631

Two lines I and J do not cross the precondition is a[i]<a[j], that is, ascending sub-sequence. Using binary search +lis, the time complexity is O (n^2), the specific explanation is described in the Challenge Program design Competition 2.3 recorded results in the use of "dynamic planning" P65

1#include <iostream>2#include <cstdio>3 using namespacestd;4 Const intmaxn=40000+5;5 Const intinf=1000000+5;6 intA[MAXN];7 intDP[MAXN];//Dp[i] Indicates the minimum value of the lowest element of the ascending subsequence of the length of i+18 intN,m,ans,l,r;9 Ten intSearchintk) One { A     intUl=l,ur=R; -      while(ur-ul>1) -     { the         intMid= (Ur+ul)/2; -         if(dp[mid]>=k) ur=mid; -             ElseUl=mid; -     } +     returnur; - } +  A intMain () at { -scanf"%d",&m); -      for(intKase=0; kase<m;kase++) -     { -scanf"%d",&n); -          for(intI=0; i<n;i++) in         { -scanf"%d",&a[i]); todp[i]=INF; +         } -l=-1; theR=0; *          for(intI=0; i<n;i++) $         {Panax Notoginseng             intpos=search (a[i]); -dp[pos]=A[i]; the             if(pos==r) r++; +         } Acout<<r<<Endl; the     } +     return 0;  -}

Dynamic planning + binary lookup poj2533&poj1631 longest ascending subsequence (LIS)