Bzoj 2209 [Jsoi2011] bracket sequence

Source: Internet
Author: User

Chain of title:

http://www.lydsy.com/JudgeOnline/problem.php?id=2209
The following:splayIt's a good question, but I'm sick of the retarded.  first it's easy to see that there's no matching parentheses at the end)))) . ((((...that is, the left side is the right parenthesis (set the number of NR), the right is to do parentheses (set the number of NL)the answer is ⌈nl÷2⌉+⌈nr÷2⌉ (⌈⌉: Rounding up) If you consider ' (' as 1, ') ' as-1,so in this sequence that contains only 1 and-1, the opposite number of the minimum prefix is equal to NR, the suffix maximum is equal to NL  so for the inquiry operation , record in splay:Key[x] (node x is 1 or-1)sum[x] (sum of x subtree corresponding to the interval)Pmn[x] (the minimum prefix value of the x subtree corresponding to the interval),Smx[x] (the maximum number of suffixes of the x subtree corresponding to the interval).  for A second reversal operation ,can be seen, just the corresponding interval of 1→-1,-1→1,so two more things to maintain.Pmx[x] (the maximum value of the prefix of the x subtree corresponding to the interval),smn[x] (the suffix of the x subtree corresponds to the minimum value). then take the key[x],sum[x],pmn[x],pmx[x],smn[x],smx[x of the record] all back (multiply by 1),and Swap (pmn[x],pmx[x]), swap (smn[x],smx[x]) (because it's reversed)just a lazy sign, please. for Third rollover Operation,Just swap the left and right subtrees,and Swap (pmx[x],smx[x]), swap (pmn[x],smn[x]) (just reverse the sequence, so swap the prefix information)then make a lazy mark.  because the lazy tag does not have a succession of influence, so lazy decentralized when you can put whichever first. Code:
#include <cstdio> #include <cstring> #include <iostream> #define MAXN 100500using namespace Std;int n,m ; struct Spt{int pmx[maxn],pmn[maxn],smx[maxn],smn[maxn],sum[maxn],key[maxn];int CH[MAXN][2],SIZ[MAXN],FA[MAXN], Lazy[maxn],rt;void Reverse (int x) {sum[x]*=-1; key[x]*=-1;pmx[x]*=-1; pmn[x]*=-1; swap (pmx[x],pmn[x]); smx[x]*=-1; SMN [X]*=-1; Swap (smx[x],smn[x]);} void Flip (int x) {swap (pmx[x],smx[x]), swap (pmn[x],smn[x]), swap (ch[x][0],ch[x][1]);} void pushup (int x) {siz[x]=siz[ch[x][0]]+1+siz[ch[x][1]];sum[x]=sum[ch[x][0]]+key[x]+sum[ch[x][1]];p Mx[x]=max (pmx [Ch[x][0]],sum[ch[x][0]]+key[x]+pmx[ch[x][1]]);p mn[x]=min (pmn[ch[x][0]],sum[ch[x][0]]+key[x]+pmn[ch[x][1]); Smx[x]=max (Smx[ch[x][1]],sum[ch[x][1]]+key[x]+smx[ch[x][0]); Smn[x]=min (SMN[CH[X][1]],SUM[CH[X][1]]+KEY[X]+SMN [Ch[x][0]]);} void pushdown (int x) {if (lazy[x]&1) {Reverse (ch[x][0]); lazy[ch[x][0]]^=1; Reverse (ch[x][1]); Lazy[ch[x][1]]^=1;lazy[x]^=1;} if (lazy[x]&2) {Flip (ch[x][0]); lazy[ch[x][0]]^=2; Flip (ch[x][1]); Lazy[ch[x][1]]^=2;lazy[x]^=2;}} void Rotate (int x,int &k) {static int y,z,l,r;y=fa[x]; z=fa[y];l=ch[y][0]!=x; r=l^1;if (!z) K=x;else ch[z][ch[z][0]!= Y]=x;fa[ch[x][r]]=y; Fa[y]=x; FA[X]=Z;CH[Y][L]=CH[X][R]; Ch[x][r]=y; Pushup (y);} void splay (int x,int &k) {static int y,z;while (X!=K) {y=fa[x]; Z=fa[y];if (y!=k) (ch[z][0]!=y) ^ (ch[y][0]!=x)? Rotate (x,k): Rotate (y,k); Rotate (x,k);} Pushup (x);} int find (int x,int num) {if (lazy[x]) pushdown (x), if (Num<=siz[ch[x][0]]) return find (ch[x][0],num); else if (num==siz[ CH[X][0]]+1) return X;else return find (ch[x][1],num-siz[ch[x][0]]-1);} int Split (int l,int r) {static int dl,dr;dl=find (RT,L); Dr=find (rt,r+2); Splay (DL,RT); Splay (dr,ch[dl][1]); return ch[dr][0];} void Modify (int l,int r,int type) {static int p;p=split (L,R), if (type==1) Reverse (p), Else Flip (p); lazy[p]^=type; Pushup (Fa[p]); Pushup (Fa[fa[p]);} void Build (int &x,int dad,int l,int r) {static char c;if (L&GT;R) return;x= (l+r) >>1; Fa[x]=dad; Build (ch[x][0],x,l,x-1); scanf ("%c", &c); key[x]= (c== ' ('? 1:-1); Build (CH[X][1],X,X+1,R); Pushup (x);} void Borderbuild () {rt=n+1;key[n+1]=0; key[n+2]=0;ch[n+1][1]=n+2; fa[n+2]=n+1; Build (Ch[n+2][0],n+2,1,n); Pushup (n+2); Pushup (n+1);} int Query (int l,int r) {static int p,ans,nl,nr;p=split (L,R); nl=-pmn[p]; nr=smx[p]; Ans= (nl+1)/2+ (nr+1)/2;return ANS;}} Dt;int Main () {freopen ("/home/noilinux/documents/Module Learning/2209.in", "R", stdin), scanf ("%d%d", &n,&m);D T. Borderbuild (); for (int i=1,c,l,r;i<=m;i++) {scanf ("%d%d%d", &c,&l,&r), if (c==0) printf ("%d\n", DT. Query (l,r)); else DT. Modify (l,r,c);} return 0;}

  

 

Bzoj 2209 [Jsoi2011] bracket sequence

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.