hdu--3308 LCIs (segment tree + interval merge)

Description

Given n integers.
You have both operations:
U A B:replace The Ath number by B. (index counting from 0)
Q A b:output The length of the longest consecutive increasing subsequence (LCIs) in [A, B].

Input

T in the first line, indicating the case number.
Each case is starts with a integers n, m (0<n,m<=10 5).
The next line has n integers (0<=val<=10 5).
The next m lines each have an operation:
U A B (0<=a,n, 0<=b=10 5)
OR
Q A B (0<=a<=b< N).

Output

For each Q, output the answer.

Sample Input

107 7 3 3 5 9 9 8 1 8 Q 6 6U 3 4Q 0 1Q 0 5Q 4 7Q 3 5Q 0 2Q 4 6U 6 10Q 0 9

Sample Output

11423125
Test instructions: Give a column n number, perform m operations, change the value of a to B, query the length of the longest continuous ascending subsequence between [A, b], and output the result for each query
Idea: Use a segment tree to save the maximum length of the current node and the length of the left start and end of the right, and then update up. The query is the same as the query in the line tree, which is to pay attention to the updates to the left and right boundaries when querying.
AC Code:

1#include <iostream>2#include <fstream>3#include <iomanip>4#include <cstdio>5#include <cstring>6#include <algorithm>7#include <cstdlib>8#include <cmath>9#include <Set>Ten#include <map> One#include <queue> A#include <stack> -#include <string> -#include <vector> the#include <ctime> -#include <sstream> -#include <cassert> - using namespacestd; + Const intmaxn=100005; - structNote + { A     intl,r,lm,rm,m; at} a[maxn<<2]; - intP[MAXN]; - intBuildintLintRintk) - { -A[k].l=l; -A[k].r=R; ina[k].lm=a[k].rm=a[k].m=0; -     if(l==R) to     { +         return 0; -     } the     intMid= (L+R)/2; *Build (l,mid,k*2); $Build (mid+1, r,k*2+1);Panax Notoginseng     return 0; - } the intPushup (intk) + { A     if(p[a[k*2].r]>=p[a[k*2+1].L]) the     { +A[k].m=max (a[k*2].m,a[k*2+1].m); -a[k].lm=a[k*2].lm; $a[k].rm=a[k*2+1].rm; $     } -     Else -     { theA[k].m=max (Max (a[k*2].rm+a[k*2+1].lm,a[k*2].M), a[k*2+1].m); -         if(a[k*2].lm== (a[k*2].r-a[k*2].l+1))Wuyia[k].lm=a[k*2].lm+a[k*2+1].lm; the         Else -a[k].lm=a[k*2].lm; Wu         if(a[k*2+1].rm== (a[k*2+1].r-a[k*2+1].l+1)) -a[k].rm=a[k*2].rm+a[k*2+1].rm; About         Else $a[k].rm=a[k*2+1].rm; -     } -     return 0; - } A intInsintNintDintk) + { the     if(a[k].l==a[k].r&&a[k].l==d) -     { $a[k].rm=a[k].lm=a[k].m=1; the         return 0; the     } the     if(a[k*2].r>=d) Ins (n,d,k*2); the     Else if(a[k*2+1].l<=d) Ins (n,d,k*2+1); - Pushup (k); in     return 0; the } the intSeaintLintRintk) About { the     if(l==a[k].l&&a[k].r==R) the     { the         returna[k].m; +     } -     if(a[k*2].R&GT;=R)returnSea (l,r,k*2); the     Else if(a[k*2+1].L&LT;=L)returnSea (l,r,k*2+1);Bayi     Else the     { the         intT1=sea (l,a[k*2].r,k*2); -         intT2=sea (a[k*2+1].l,r,k*2+1); -         intans=Max (T1,T2); the         intsum=0; the         if(p[a[k*2].r]<p[a[k*2+1].L]) theSum=min (a[k*2].r-l+1, a[k*2].RM) +min (r-a[k*2].r,a[k*2+1].lm); the         returnMax (ans,sum); -     } the } the intMain () the {94     intn,m,t,c,b; thescanf"%d",&t); the     { the          for(intkk=1; kk<=t; kk++)98         { Aboutscanf"%d%d",&n,&m); -Build1N1);101              for(intI=1; i<=n; i++)102             {103scanf"%d",&c);104p[i]=C; theIns (C,i,1);106             }107             Chars[Ten];108              for(intI=0; i<m; i++)109             { thescanf"%s", s);111                 if(s[0]=='U') the                 {113scanf"%d%d",&c,&b); thep[c+1]=b; theINS (b,c+1,1); the                 }117                 Else118                 {119scanf"%d%d",&c,&b); -printf"%d\n", Sea (c+1, B +1,1));121                 }122             }123         }124     } the     return 0;126}

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hdu--3308 LCIs (segment tree + interval merge)