HDU 1950 Bridging signals (LIS, dichotomy, O (NLOGN))

Test instructions

Given a sequence of numbers, it is required to find the LIS and output its length.

Ideas:

Sweep over + two points, complexity O (NLOGN), Spatial complexity O (n).

Method: Add an array, using d[i] to represent the last element of the increment subsequence of length I, and the element always remains the current minimum. Initialize D[1]=a[i], the length of the current LIS is len=1. From 2 to N, if A[i]>d[len], then d[++len]=a[i], otherwise, find in the array D a[i] should be inserted in place, instead of the first larger than the number, such as d[k]<a[i]<=d[k+1], directly a[i] Instead of d[k+1]. After completion len is the length of the LIS.

1#include <bits/stdc++.h>2#include <iostream>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <map>7#include <algorithm>8#include <vector>9#include <iostream>Ten #definePII pair<int,int> One #defineINF 2147483647 A #defineLL unsigned long Long - using namespacestd; - Const DoublePI = ACOs (-1.0); the Const intn=40100; - Const intmod=1e9+7; - intA[n], d[n]; - int* Lower_ (int*s,int*e,intVal)//binary Find value, return subscript + { -     intL=0, r=e-s-1, mid; +      while(l<R) A     { atmid=r-(r-l+1)/2;//guaranteed at least 1 reduction -         if(S[mid]<val) L=mid+1;//at least 1 increase -         ElseR=mid; -     } -     return&S[r]; - } in  - intMain () to { +Freopen ("Input.txt","R", stdin); -     intT, N, Len; theCin>>T; *      while(t--) $     {Panax Notoginsengscanf"%d",&n); -          for(intI=1; i<=n; i++) scanf ("%d",&a[i]); thelen=1; +d[len]=a[1]; A  the          for(intI=2; i<=n; i++ ) +         { -             if(A[i]>d[len]) d[++len]=A[i]; $             Else*lower_ (d+1, d+len+1, a[i]) =A[i]; $             //else *lower_bound (D+1,d+len+1,a[i]) =a[i]; the previous line of code can be replaced by this line -         } -printf"%d\n", Len); the     } -     return 0;Wuyi}

AC Code

HDU 1950 Bridging signals (LIS, dichotomy, O (NLOGN))