Bzoj 2209: [Jsoi2011] bracket sequence [splay brackets]

2209: [Jsoi2011] Bracket sequence time limit:20 Sec Memory limit:259 MB
submit:1111 solved:541
[Submit] [Status] [Discuss] The first line of the Descriptioninput input data contains two integers n and q, each representing the length of the bracket sequence and the number of operations. The second line contains a sequence of parentheses with a length of N. Next Q line, each line of three integers t, x and Y, respectively, indicating the type of operation, the starting position of the operation and the end position of the operation, the input data guaranteed x is not less than Y. Where t=0 indicates an inquiry operation, T=1 represents a reversal operation, and t=2 represents a rollover operation. Output for each query operation, a line is shown that indicates the minimum number of changes required to modify the sequence of parentheses to match the subsequence. Sample Input6 3
)(())(
0 1 6
0 1 4
0 3 4Sample Output2
2
0
HINT

100% of data meet n,q not more than 10^5
。

Source

First round

Background: The key lock room in the computer can only come to the electronic reading room, in the cold humid air unexpectedly 1 A is too moved ( although not entirely self-thought out )

Parentheses question a common technique is (-1) +1

The last to be modified)) (the number of ((this way)) is the minimum prefix and ((

So the number of changes is (lmn+1)/2+ (rmx+1)/2

Information merging and dynamic maximal sub-sequences the problem is very similar.

How is information maintained after flipping and reversing?

Flip: Just swap L. and R.. Just fine.

Invert: all. mx. MN must be negative, then swap (LMX,LMN) swap (RMX,RMN)

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespaceStd;typedefLong Longll;#defineLC T[x].ch[0]#defineRC T[x].ch[1]#definePA T[X].FAConst intn=1e5+5, inf=1e9;inlineintRead () {CharC=getchar ();intx=0, f=1;  while(c<'0'|| C>'9'){if(c=='-') f=-1; C=GetChar ();}  while(c>='0'&&c<='9') {x=x*Ten+c-'0'; C=GetChar ();} returnx*F;}intN,q,l,r,a[n];Chars[n],op[ -];structnode{intch[2],FA,SIZE,V,SUM,LMX,LMN,RMX,RMN,REV,FLP; Node (): FA (0) {ch[0]=ch[1]=0;}} T[n];intRoot;inlineintWhintx) {returnt[pa].ch[1]==x;} InlinevoidUpdateintx) {    if(!x)return; T[x].size=t[lc].size+t[rc].size+1; T[x].sum=t[lc].sum+t[rc].sum+t[x].v; T[X].LMX=max (T[lc].lmx,max (t[lc].sum+t[x].v,t[lc].sum+t[x].v+t[rc].lmx)); T[X].LMN=min (T[lc].lmn,min (t[lc].sum+t[x].v,t[lc].sum+t[x].v+t[rc].lmn)); T[x].rmx=max (T[rc].rmx,max (t[rc].sum+t[x].v,t[rc].sum+t[x].v+t[lc].rmx)); T[X].RMN=min (T[rc].rmn,min (t[rc].sum+t[x].v,t[rc].sum+t[x].v+t[lc].rmn)); }inlinevoidRever (intx) {T[x].rev^=1;    Swap (LC,RC);    Swap (T[X].LMX,T[X].RMX); Swap (T[X].LMN,T[X].RMN);} InlinevoidFlipintx) {T[X].FLP^=1; T[x].sum=-t[x].sum;t[x].v=-t[x].v; T[X].LMX=-t[x].lmx;t[x].lmn=-T[X].LMN;    Swap (T[X].LMX,T[X].LMN); T[x].rmx=-t[x].rmx;t[x].rmn=-T[X].RMN; Swap (T[X].RMX,T[X].RMN);} InlinevoidPushdown (intx) {    if(T[x].rev) {if(LC) Rever (LC); if(RC) rever (RC); T[x].rev=0; }    if(T[X].FLP) {if(LC) Flip (LC); if(RC) flip (RC); T[X].FLP=0; }}inlinevoidRotateintx) {    intF=t[x].fa,g=t[f].fa,c=WH (x); if(g) T[g].ch[wh (f)]=x;t[x].fa=G; T[F].CH[C]=t[x].ch[c^1];t[t[f].ch[c]].fa=F; T[x].ch[c^1]=f;t[f].fa=x; Update (f); update (x);} InlinevoidSplay (intXinttar) {     for(;p a!=tar;rotate (x))if(T[pa].fa!=tar) Rotate (WH (x) ==WH (PA)?pa:x); if(tar==0) root=x;}intBuildintLintRintf) {//printf ("Build%d%d%d\n", l,r,f);    if(L>r)return 0; intx= (l+r) >>1; LC=build (l,x-1, x); Rc=build (x+1, r,x); T[X].FA=F; T[x].rev=t[x].flp=0;//printf ("Al%d\n", A[l]);T[X].V=A[X];//Not needUpdate (x);//printf ("Get%d%d%d%d%d%d%d%d %d\n", X,L,R,T[X].V,T[X].SUM,T[X].LMX,T[X].LMN,T[X].RMX,T[X].RMN);    returnx;} InlineintKthintK) {//printf ("kth%d\n", k);    intx=root,ls=0;  while(x) {pushdown (x); int_=ls+t[lc].size;//printf ("Size%d%d\n", x,_);        if(_<k&&k<=_+1)returnx; Else if(k<=_) x=LC; Elsels=_+1, x=RC; }    return 0;}voidQuery (intLintR) {//printf ("Query%d%d\n", l,r);    intF=kth (l); Splay (F,0); intX=kth (r+2); splay (x,f); intA=T[LC].LMX,B=T[LC].RMN;//printf ("Hi%d%d%d%d%d\n", f,x,lc,a,b);printf"%d\n", (A +1)/2+ (-b+1)/2);}voidFlip (intLintR) {    intF=kth (l); Splay (F,0); intX=kth (r+2); splay (x,f); Flip (LC); update (x); update (f);}voidRever (intLintR) {    intF=kth (l); Splay (F,0); intX=kth (r+2); splay (x,f); Rever (LC); update (x); update (f);}intMain () {//freopen ("In.txt", "R", stdin);N=read (); q=read (); scanf ("%s", s+1);  for(intI=1; i<=n;i++) a[i+1]=s[i]=='('?-1:1; //for (int i=1;i<=n+2;i++) printf ("%d", A[i]);p UTS ("");t[0].lmn=t[0].rmn=INF; t[0].lmx=t[0].rmx=-INF; Root=build (1, n+2, Root);  while(q--) {scanf ("%s", op); L=read (); r=read (); if(op==0) Query (l,r); if(op==1) Flip (l,r); if(op==2) Rever (l,r); }}

Bzoj 2209: [Jsoi2011] bracket sequence [splay brackets]