Lis Longest ascending subsequence O (nlogn) algorithm

The normal way to find the LIS is to do with DP, time complexity of O (n^2), but the face of some problems when the complexity is a bit high, to learn a NLOGN solution. The main use of two points to find, STL inside the Lower_bound can also.

Upper_bound (i) returns the last position (upper bound) where the element with the key value of I can be inserted
Lower_bound (i) returns the first position of the position where the element with the key value of I can be inserted (lower bound)

Put an O (n^2) code First

1#include <bits/stdc++.h>2 #defineFi first3 #defineSe Second4 #definePB Push_back5 #defineFio Ios::sync_with_stdio (false); Cin.tie (0);6 #definePII pair<int,int>7 #defineVI vector<int>8 #defineVC Vector<char>9 #defineMii map<int,int>Ten #defineSi (a) scanf ("%d", &a) One #defineSS (a) scanf ("%s", &a) A #defineSL (a) scanf ("%i64d", &a); - #defineSLF (a) scanf ("%lf", &a); - #defineCLEAR (A, B) memset (A,b,sizeof (a)) the #definePi ACOs (-1) -  - Const intinf=0x3f3f3f3f; - Const intn=2e5+5; +  -typedefLong Longll; +typedefDoubledb; Atypedef unsignedLong Longull; at using namespacestd; - intDp[n]; - intNum[n]; - intN; -  - voidSolve () in { -     intres=0; to      for(intI=0; i<n;i++) +     { -dp[i]=1; the          for(intj=0; j<i;j++) *         { $             if(num[j]<Num[i])Panax Notoginseng             { -                  Do[I]=max (dp[i],dp[j]+1); the             } +res=Max (res,dp[i]); A         } the     } +cout<<res<<Endl; - } $ intMain () $ { - Fio; -     intN; theCin>>N; -      for(intI=0; i<n;i++)Wuyi     { theCin>>Num[i]; -     } Wu solve (); -}

This is very good understanding, for O (NLOGN) solution I read some blog, the more important thing is to replace this operation, each time to find a more than the last value of the time to join the back, hit the small number than the last face when looking for the first one smaller than its number, replace the back one. (Chinese is not good ...) Just look at the example.)

such as 2 5 3 4 9 7 8 10 It is easy to see that its LIS length is 6, the longest is 2 3 4 7 8 10.

The concrete steps are as follows, 2 enter; 5:2 large, enter; At this time because 3:5 small, looking for, 2:3 small, 3 replace 5, 4:3, 4, 9:4, 9, 7:9,, Forward, 4:7, 7, replace 9, followed by direct access, the LIS sequence 2 3 4 7 8 10.

O (NLOGN) code is specified as follows

1#include <bits/stdc++.h>2 #defineFi first3 #defineSe Second4 #definePB Push_back5 #defineFio Ios::sync_with_stdio (false); Cin.tie (0);6 #definePII pair<int,int>7 #defineVI vector<int>8 #defineVC Vector<char>9 #defineMii map<int,int>Ten #defineSi (a) scanf ("%d", &a) One #defineSS (a) scanf ("%s", &a) A #defineSL (a) scanf ("%i64d", &a); - #defineSLF (a) scanf ("%lf", &a); - #defineCLEAR (A, B) memset (A,b,sizeof (a)) the #definePi ACOs (-1) -  - Const intinf=0x3f3f3f3f; - Const intn=2e5+5; +  -typedefLong Longll; +typedefDoubledb; Atypedef unsignedLong Longull; at using namespacestd; - intAns[n]; - intNum[n]; -  - intMain () - { in Fio; -     intN; toCin>>N; +      for(intI=1; i<=n;i++) -Cin>>Num[i]; theans[1]=num[1]; *     intlen=1; $      for(intI=2; i<=n;i++)Panax Notoginseng     { -         if(num[i]>Ans[len]) theans[++len]=Num[i]; +         Else A         { the             intPos=lower_bound (Ans,ans+len,num[i])-ans; +ans[pos]=Num[i]; -         } $     } $      for(intI=1; i<=len;i++) -     { -cout<<ans[i]<<" "; the     } -cout<<Endl;Wuyi } the  -  Wu /* - 8 About 2 5 3 4 9 7 8 $ */

Lis Longest ascending subsequence O (nlogn) algorithm