[Rope algorithm] The Persistence balancing tree in STL-rope

Simple usage:

# Include <EXT/rope> using namespace _ gnu_cxx; int A [1000]; rope <int> X (A, A + n ); rope <int> A (x); X-> at (10); X [10]; X-> push_back (X) // Add XX-> insert (Pos, x) at the end. // insert XX-> erase (Pos, x) at the POS) // Delete X from pos-> Replace (Pos, x) // replace from POS with XX-> substr (Pos, x) // extract X from POS

 

Example 1:

Ioi2012

Scrivener

Question

The design supports the following three operations:
1. t x: enter a lowercase letter X at the end of the article. (Type Operation)
2. u x: undo the last X Modification Operation. (Undo operation)
(Note that the query operation is not a modification operation)
3. q X: Ask the X letter in the current article and output it. (Query Operation)

Operand n <= 100000 online Algorithm

CLJ says this is a silly question of rope ......

 

#include<cstdio>#include<cstring>#include<cctype>#include<iostream>#include<algorithm>#include<ext/rope>using namespace std;using namespace __gnu_cxx;const int maxn=1e5+10;rope<char> *his[maxn];int n;int d[maxn];inline int lowbit(int x){ return x&-x; }inline void updata(int x){ while(x<=n){ d[x]++; x+=lowbit(x); } }inline int get(int x){ int res=0; while(x){ res+=d[x]; x-=lowbit(x); }return res; }inline char getC(){ char ch=getchar(); while(!isalpha(ch))ch=getchar(); return ch; }inline int getint(){    int res=0; char ch,ok=0;    while(ch=getchar())    {        if(isdigit(ch)){ res*=10;res+=ch-‘0‘;ok=1; }        else if(ok)break;    }    return res;}void deb(rope<char> s){    for(int i=0;i<s.length();i++)        cout<<s[i];puts("");}int main(){//    freopen("type.in","r",stdin);//    freopen("type.out","w",stdout);    n=getint();    his[0]=new rope<char>();    for(int i=1;i<=n;i++)    {        his[i]=new rope<char>(*his[i-1]);        // deb(*his[i]);        char opt=getC();        if(opt==‘T‘)        {            char x=getC();            his[i]->push_back(x);            updata(i);        }        else if(opt==‘U‘)        {            updata(i);            int x=getint();            int l=1,r=i,mid,now=get(i);            while(l<r)            {                mid=(l+r)>>1;                if(now-get(mid)>x) l=mid+1;                else r=mid;            }            his[i]=his[l-1];        }        else if(opt==‘Q‘)        {            int x=getint()-1;            putchar(his[i]->at(x));            putchar(‘\n‘);        } // deb(*his[i]);    }    return 0;}

 

Among them, the operation to achieve persistence is: His [I] = new rope <char> (* his [I-1]);

It can implement the copy History version of O (1). Because rope's bottom layer is a red/black tree, you only need to copy the root pointer during copy.

One-click persistence ......

 

Example 2:

Bzoj 3673:Persistent and query set by zky

N sets, M operations
Operation:
1 a B Merge A and B in the Set
2 k returns to the status after the K operation (the query is counted as an operation)
3 a B: Check whether a and B belong to the same set. If yes, output 1. Otherwise, output 0.

0 <n, m <= 2*10 ^ 4

Analysis: You can directly persistently query the FA array of the set.

Accode:

#include <cstdio>#include <ext/rope>using namespace std;using namespace __gnu_cxx;const int maxm = 20010;rope<int> *rp[maxm];int find(int i,int x){    if(rp[i]->at(x) == x)   return x;    int f = find(i,rp[i]->at(x));    if(f == rp[i]->at(x))   return f;    rp[i]->replace(x,f);    return rp[i]->at(x);}inline void merge(int i,int x,int y){    x = find(i,x),y = find(i,y);    if(x != y)  rp[i]->replace(y,x);}int a[maxm],lastans;int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i = 1;i <= n;i ++)  a[i] = i;    rp[0] = new rope<int> (a,a + n + 1);    for(int i = 1;i <= m;i ++)    {        rp[i] = new rope<int> (*rp[i - 1]);        int opt;    scanf("%d",&opt);        if(opt == 1)        {            int a,b;    scanf("%d%d",&a,&b);            merge(i,a/* ^ lastans*/,b/* ^ lastans*/);        }        else if(opt == 2)        {            int k;  scanf("%d",&k);            rp[i] = rp[k/* ^ lastans*/];        }        else        {            int a,b;    scanf("%d%d",&a,&b);            printf("%d\n",lastans = (find(i,a/* ^ lastans*/) == find(i,b/* ^ lastans*/)));        }    }    return 0;}

 

Example 3:

Ahoi2006 text editor Editor

Question

Design Data Structure support

Insert and delete reverse string

Analysis:

Because of the underlying implementation of rope, insert, erase, and get are all logn

You can't flip it. You can't mark it manually !!

What should I do?

A: At the same time, we maintain one, one, and two ropes ...... Invert: swap two substrings ...... Orz ......

Interval Cyclic Displacement? Simple. Just split them into multiple sub-databases and concatenate them ......

Range A to B to C to D ...... Z changed to? Er ...... Maintain 26 ropes?

Interval and? It's a live line tree.

Interval kth? Sorry, numeric-related operations rope does not support ......

5555 the maintenance series can only be written by yourself ......

Accode:

#include <cstdio> #include <ext/rope> #include <iostream> #include <algorithm> using namespace std; using namespace __gnu_cxx; crope a,b,tmp; char s[10]; int now,n,len,size; char str[2000000],rstr[2000000]; int main(){     scanf("%d",&n);     while(n--)    {         scanf("%s",s);         switch(s[0])        {             case ‘M‘:            {                scanf("%d",&now);                break;            }             case ‘P‘:            {                now--;                break;            }             case ‘N‘:            {                now++;                break;            }             case ‘G‘:            {                putchar(a[now]);                putchar(‘\n‘);                break;            }             case ‘I‘:            {                 scanf("%d",&size);                 len=a.length();                 for(int i=0;i<size;i++)                {                     do                    {                        str[i]=getchar();                    }                     while(str[i]==‘\n‘);                     rstr[size-i-1]=str[i];                 }                 rstr[size]=str[size]=‘\0‘;                 a.insert(now,str);                 b.insert(len-now,rstr);                 break;             }                           case ‘D‘:            {                 scanf("%d",&size);                 len=a.length();                 a.erase(now,size);                 b.erase(len-now-size,size);                 break;             }             case ‘R‘:            {                 scanf("%d",&size);                 len=a.length();                 tmp=a.substr(now,size);                 a=a.substr(0,now) + b.substr(len-now-size,size) + a.substr(now+size,len-now-size);                 b=b.substr(0,len-now-size)+tmp+b.substr(len-now,now);                 break;             }         }     }     return 0; }

 

 

End