[JSOI2008] Mars people

This topic I successfully burst zero,\ (\color{blue}{tttttle}\) the whole point ... I'm really good.

\ (\color{red}{description}\)

The Martians have recently studied an operation that asks for a common prefix of two suffixes of a string. For example, there is a string like this:\ (madamimadam\),
We label each character of this string: ordinal: 1 2 3 4 5 6 7 8 9 10 11 character M A d a m i m a D a m now,
The Martians define a function \ (LCQ (x, y) \), which is a string of characters starting with the first x character of the substring, and a string beginning with the first character of the y\
The length of the public prefix of two strings. For example,\ (LCQ (1, 7) = 5\), \ (LCQ (2) = 1, LCQ (4, 7) = 0\) in the process of studying the \ (lcq\) function
, the Martians discovered such an association: if all suffixes of the string are ordered, the value of the \ (lcq\) function can be quickly calculated;
If you find the value of the LCQ function, you can also quickly sequence the suffix of the string. Although the Martians were clever enough to find a quick way to get the \ (lcq\) function
Algorithm, but the people who do not willingly concede to the earth have a problem for the Martians: in the lcq\ function, you can also change the string itself. Specifically
, you can change the value of one of the characters in the string, or you can insert a character at one point in the string. Earth people want to test, in so
Complex question, whether the Martians can also be able to quickly find the value of the \ (lcq\) function.

\ (\color{red}{input}\)

The first line gives the initial string. The second line is a non-negative integer\ (m\)That represents the number of actions. The next\ (m\)Row, each line describes an action. Exercise
There are 3 types, as shown below
\ (1\)Ask. Grammar:\ (qxy\)，\ (x,y\)are positive integers. Function: Calculation\ (LCQ (x, y) \)Limit: 1<=x,y<= the current string length.
\ (2\)Modify. Grammar:\ (rxd\)，\ (x\)is a positive integer,\ (d\)is a character. Function: The string in the\ (x\)Number modified to character\ (d\)。 Limit:\ (x\)No more than the current word
The length of the character string.
\ (3\), insert: Syntax:\ (ixd\)，\ (x\)is a non-negative integer,\ (d\)is a character. Function: in string section\ (x\)Characters and then insert the character D, if\ (x=0\), then in the word
Inserted at the beginning of the symbol string. Limit:\ (x\)Does not exceed the current string length

\ (\color{red}{output}\)

For each query in the input file, you should output the corresponding answer. One answer line.

\ (\color{red}{sample \ \ input}\)

Madamimadam
7
Q 1 7
Q 4 8
Q 10 11
R 3 A
Q 1 7
I Ten A
Q 2 11

\ (\color{red}{sample \ \ output}\)

5
1
0
2
1

\ (\color{red}{hint}\)

1. All strings are composed of lowercase letters from beginning to finish.

2,\ (m<=150,000\)

3, String length L meet l<=100,000 from beginning to finish

4, the number of inquiry operation is not more than 10,000.

For the 1th, 2 data, the string length does not exceed 1,000 from beginning to the other

For the 3,4,5 data, there is no insert operation.

\ (\color{purple}{half-solution}\)

Well, compare the record, I will only temporarily tle\ version of violence, first posted to record my glorious history .

Insert is inserted, modification is modified, but because in fact I was only half- splay\ , so now the interval \ (splay\) at a loss (otz\)...

Well, the query, then, is to constantly query their parent node, or right child node, but because you \ (splay\) words will make the sequence unordered, so it is not possible to do.

So in fact, you see, this is actually a (wa\) but first \ (tle\) algorithm.

hahaha I also debugged for half a day.

That's funny.

#include <iostream>#include <cstdio>#define MAXN 250001#define IL inlineusing namespaceStdstructchars{unsigned intson[2],f,sub;CharV;} S[MAXN];CharT,BASE[MAXN],CC;unsigned intK,res;unsigned intXx,yy,m,a,b,i,qwq,root,wz,now,fnow,ffnow;ilBOOLWunsigned intx) {returnx==s[s[x].f].son[1];} IlvoidPUSH_UP (unsigned intx) {if(x) {s[x].sub=1;if(s[x].son[1]) s[x].sub+=s[s[x].son[1]].sub;if(s[x].son[0]) s[x].sub+=s[s[x].son[0]].sub; }}ilvoidRotateunsigned intx) {fnow=s[x].f,ffnow=s[fnow].f;BOOLWw=w (x); s[fnow].son[ww]=s[x].son[ww^1];    S[s[fnow].son[ww]].f=fnow;    S[fnow].f=x;    S[x].f=ffnow; s[x].son[ww^1]=fnow;if(Ffnow) {s[ffnow].son[s[ffnow].son[1]==fnow]=x;    } push_up (Fnow); PUSH_UP (x);}voidSplay (unsigned intXunsigned intGoal) { for(QWQ; (QWQ=S[X].F)!=goal;rotate (x)) {if(S[qwq].f!=goal)        {Rotate (w (x) ==w (QWQ) qwq:x); }    }if(goal==0) {root=x; }}voidInsertunsigned intPosCharc) {if(!WZ)        {wz++;        ROOT=WZ; s[wz].f=s[wz].son[1]=s[wz].son[0]=0;        s[wz].sub++; S[wz].v=c;return; }if(!pos)        {wz++; s[wz].son[1]=root,s[wz].v=c;        S[ROOT].F=WZ;        ROOT=WZ; PUSH_UP (WZ);return; } now=root; while(1){if(s[now].son[0]&&pos<=s[s[now].son[0]].sub) now=s[now].son[0];Else{unsigned intTemp= (s[now].son[0]?s[s[now].son[0]].sub:0)+1;if(pos<=temp) Break;            Pos-=temp; now=s[now].son[1];    }} wz++; s[wz].son[1]=s[now].son[1]; s[s[now].son[1]].F=WZ; s[now].son[1]=WZ;    S[wz].f=now;    S[wz].v=c;    PUSH_UP (WZ); PUSH_UP (now);//splay (wz,0);}ilvoidUpdateunsigned intPosCharc) {now=root; while(1){if(s[now].son[0]&&pos<=s[s[now].son[0]].sub) now=s[now].son[0];Else{unsigned intTemp= (s[now].son[0]?s[s[now].son[0]].sub:0)+1;if(pos<=temp) Break;            Pos-=temp; now=s[now].son[1]; }} s[now].v=c;//Splay (now,0);}ilunsigned intFindunsigned intSxunsigned intSY) {xx=root,res=0, Yy=root; while(1){if(s[xx].son[0]&&sx<=s[s[xx].son[0]].sub) xx=s[xx].son[0];Else{unsigned intTemp= (s[xx].son[0]?s[s[xx].son[0]].sub:0)+1;if(sx<=temp) Break;            Sx-=temp; xx=s[xx].son[1]; }    } while(1){if(s[yy].son[0]&&sy<=s[s[yy].son[0]].sub) yy=s[yy].son[0];Else{unsigned intTemp= (s[yy].son[0]?s[s[yy].son[0]].sub:0)+1;if(sy<=temp) Break;            Sy-=temp; yy=s[yy].son[1]; }    } while(1){if(S[YY].V==S[XX].V) res++; yy=s[yy].son[1]; xx=s[xx].son[1];if(!yy| |! xx| | S[YY].V!=S[XX].V) {returnRes }}}ilunsigned intQread () {k=0, Cc=getchar (); while(!isdigit (CC))    {Cc=getchar (); } while(IsDigit (CC)) {k= (k<<1) + (k<<3) +CC-48;    Cc=getchar (); }returnK;}intMain () {scanf ("%s", base); M=qread (); for(i=0; base[i];i++) {insert (i,base[i]); } for(i=1; i<=m;i++) {scanf ("%c", &t);if(t==' I ') {a=qread (); scanf"%c", &t); Insert (A,T);Continue; }Else if(t==' R ') {a=qread (); scanf"%c", &t); Update (A,T);Continue;        } a=qread ();        B=qread (); printf"%d\n", find (b)); }return 0;}

The above has been hung on the stigma of history

[JSOI2008] Mars people