HDU 5489 removed Interval (LIS variant)

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

Give you the number of n, to delete the consecutive L, ask you to delete after the maximum number of LIS?

We first preprocess the LIS, which begins with the I subscript, into the array.

You can then enumerate the ranges that are long as l, each time you move, add one to the left, and delete one on the right.

Longest ascending subsequence length = the longest ascending subsequence of the window to the right of the first element to the right and the longest ascending sub-sequence of the left side of the window that is smaller than the window to the right of the first element.

For example 1 2 [4 3] 2 3 5, LIS = 3 + 1 = 4

The LIS that begins with I, as long as the longest descending subsequence or inverted a[i] becomes negative, the longest ascending subsequence can be obtained.

The code is a bit messy.

1 //#pragma COMMENT (linker, "/stack:102400000, 102400000")2#include <algorithm>3#include <iostream>4#include <cstdlib>5#include <cstring>6#include <cstdio>7#include <vector>8#include <cmath>9#include <ctime>Ten#include <list> One#include <Set> A#include <map> - using namespacestd; -typedefLong LongLL; thetypedef pair <int,int>P; - Const intN = 1e5 +5; - intDp1[n], Dp2[n], a[n], inf = 1e9 +7; - intY[n], x[n];//Y[i] indicates that lis,x[i, which begins with the I subscript), represents the LIS ending with I +  - intMain () + { A     intt, N, M; atscanf"%d", &t); -      for(intCA =1; CA <= T; ++CA) { -scanf"%d%d", &n, &m); -          for(inti =1; I <= N; ++i) { -scanf"%d", A +i); -Dp1[i] = Dp2[i] =inf; in         } -dp2[0] = dp1[0] =inf; to         intAns =0;  +          for(inti = n; I >=1; -I.) {//The lis that begins with I, seeking to change negative -             intpos = Lower_bound (DP2, DP2 + N,-a[i])-DP2; theY[i] = pos +1; *             //cout << y[i] << Endl; $Dp2[pos] =-A[i];Panax Notoginseng             if(i = = m +1) { -Ans = lower_bound (DP2, DP2 + N, INF)-DP2; the             } +         } A         //cout << ans << endl; the          for(inti =1; I <= N; ++i) { +             intpos = Lower_bound (DP1, DP1 + N, a[i])-DP1; -X[i] = pos +1; $Dp1[pos] =A[i]; $             if(i = = N-m) { -Ans = max (int(Lower_bound (DP1, DP1 + N, INF)-dp1), ans); -                  Break; the             } -         }Wuyi         //cout << ans << endl; theprintf"Case #%d:", CA); -dp1[0] =inf; Wu          for(inti =1; I <= N; ++i) { -Dp1[i] =inf; About         } $          for(inti = m +1; I <= N; ++i) { -             intpos = Lower_bound (DP1, DP1 + N, a[i]-1)-DP1;//the left side of the sliding window is just less than the first number of LIS on the right. -ans = max (ans, y[i] + (A[i]-1= = A[pos]? POS +1: POS)); -*lower_bound (DP1, DP1 + N, a[i-m]) = A[i-m]; A         } +printf"%d\n", ans); the     } -     return 0; $}

HDU 5489 removed Interval (LIS variant)