"Dichotomy", "Longest ascending subsequence" HDU 5489 removed Interval (acm/icpc Asia regional Hefei Online)

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=5489

Main topic:

A sequence of N (n<=100000) that removes the number of adjacent L (minus the entire interval), making the longest ascending subsequence (LIS) the longest remaining.

Topic Ideas:

"Dichotomy" "Longest ascending subsequence"

First, assuming that the number of [i,i+m-1] L is removed, the remaining LIS length is Max (the last Lis length at the end of the I left is less than a[i+m] +a[i+m] start to the last Lis length).

So we can know the LIS that starts at the end of each number by f[i the length of the longest descent subsequence from N to 1.

Then, from the beginning to do the LIS, while the second in the number of the first I find the last one is smaller than a[i+m] the longest lis length, update the answer.

1 //2 //by Coolxxx3 //#include <bits/stdc++.h>4#include <iostream>5#include <algorithm>6#include <string>7#include <iomanip>8#include <map>9#include <stack>Ten#include <queue> One#include <Set> A#include <bitset> -#include <memory.h> -#include <time.h> the#include <stdio.h> -#include <stdlib.h> -#include <string.h> - //#include <stdbool.h> +#include <math.h> - #defineMin (a) < (b) ( A):(B)) + #defineMax (a) (a) > (b)? ( A):(B)) A #defineABS (a) ((a) >0? ( A):(-(a))) at #defineLowbit (a) (a& (a)) - #defineSqr (a) ((a) * (a)) - #defineSwap (a) (a) ^= (b), (b) ^= (a), (a) ^= (b)) - #defineMem (A, B) memset (A,b,sizeof (a)) - #defineEPS (1E-8) - #defineJ 10000 in #defineMoD 1000000007 - #defineMAX 0x7f7f7f7f to #definePI 3.14159265358979323 + #defineN 100004 - using namespacestd; thetypedefLong LongLL; * intCas,cass; $ intN,m,lll,ans;Panax Notoginseng LL Aans; - intA[n],f[n],q[n]; the voidGETF () + { A     intI,l,r,mid; theLll=0; +      for(i=n;i;i--) -     { $L=0, r=lll; $          while(l<R) -         { -Mid= (l+r+1) >>1; the             if(Q[mid]>a[i]) l=mid; -             Elser=mid-1;Wuyi         } theq[r+1]=A[i]; -f[i]=r+1; WuLll=max (lll,r+1); -     } About } $ voidWork () - { -     intI,l,r,mid; -Lll=0; A      for(i=1; i+m<=n;i++) +     { theL=0, r=lll; -          while(l<R) $         { theMid= (l+r+1) >>1; the             if(Q[mid]<a[i+m]) l=mid; the             Elser=mid-1; the         } -Ans=max (ans,r+f[i+m]); inL=0, r=lll; the          while(l<R) the         { AboutMid= (l+r+1) >>1; the             if(Q[mid]<a[i]) l=mid; the             Elser=mid-1; the         } +q[r+1]=A[i]; -Lll=max (lll,r+1); the     }Bayians=Max (ans,lll); the } the intMain () - { - #ifndef Online_judge the //freopen ("1.txt", "R", stdin); the //freopen ("2.txt", "w", stdout); the     #endif the     inti,j,k; -      the //for (scanf ("%d", &cass); cass;cass--) the      for(SCANF ("%d", &cas), cass=1; cass<=cas;cass++) the //while (~scanf ("%s", s+1))94 //while (~scanf ("%d", &n)) the     { theans=0; theprintf"Case #%d:", Cass);98scanf"%d%d",&n,&m); About          for(i=1; i<=n;i++) scanf ("%d", A +i); - getf ();101 Work ();102printf"%d\n", ans);103     }104     return 0; the }106 /*107 //108 109 // the */

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"Dichotomy", "Longest ascending subsequence" HDU 5489 removed Interval (acm/icpc Asia regional Hefei Online)