12538 Version Controlled IDE 
Programmers use version control systems to manage les in their projects, but in these systems, versions 
are saved only when you manually submit. Can you implement an IDE that automatically saves a new 
version whenever you insert or delete a string? 
Positions in the bu er are numbered from 1 from left to right. Initially, the bu er is empty and in 
version 0. Then you can execute 3 commands (vnow means the version before executing the command, 
and L[v] means the length of bu er at version v): 
1 p s: insert string s after position p (0 p L[vnow], p = 0 means insert before the start of the 
bu er). s contains at most 1 and at most 100 letters. 
2 p c: remove c characters starting at position p (p 1; p + c L[vnow] + 1). The remaining 
charactesr (if any) will be shifted left, lling the blank 
3 v p c: print c characters starting at position p (p 1; p + c L[v] + 1), in version v (1 v 
vnow). 
The rst command is guaranteed to be command 1(insert). After executing each command 1 or 2, 
version is incremented by 1. 
Input 
There is only one test case. It begins with a single integer n (1 n 50; 000), the number of commands. 
Each of the following n lines contains a command. The total length of all inserted string will not 
exceed 1,000,000. 
Output 
Print the results of command 3, in order. The total length of all printed strings will not exceed 200,000. 
Obfuscation: 
In order to prevent you from preprocessing the command, we adopt the following obfuscation scheme: 
Each type-1 command becomes 1 p + d s 
Each type-2 command becomes 2 p + d c + d 
Each type-3 command becomes 3 v + d p + d c + d 
Where d is the number of lowercase letter `c' you printed, before processing this command. 
Before the obfuscation, the sample input would be: 
6 
1 0 abcdefgh 
2 4 3 
3 1 2 5 
3 2 2 3 
1 2 xy 
3 3 2 4 
This is the real input that your program must process when it reads the Sample Input 
below.Universidad de Valladolid OJ: 12538 { Version Controlled IDE 2/2 
Sample Input 
6 
1 0 abcdefgh 
2 4 3 
3 1 2 5 
3 3 3 4 
1 4 xy 
3 5 4 6 
Sample Output 
bcdef 
bcg 
bxyc 

維護一個文字，p位置插入串，p位置刪除長度c的串，
查詢歷史版本某個位置開始的串。強制線上
神一樣的可持久化treap

不用rotate操作
merge(a，b)：把 treap<a>與treap<b >合併成treap<c>
split(a，k)：把treap<a >的前k個變成一個treap<b>，剩下的元素變成treap<c>

研究中，程式碼是悄悄拿來參考噠

#include<cstdio>
#include<ctime>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1111111;
int dif,m,th;
char key[maxn*10];
int root[55555]={0};
char s[maxn];
int son[maxn*20][2]={0};
int size[maxn*20]={0};
int val[maxn*20]={0};
int stuck[maxn]={0};
int top=0;
inline void update(int x)
{
    if (size[son[x][0]]==0)son[x][0]=0;
    if (size[son[x][1]]==0)son[x][1]=0;
    size[x]=size[son[x][0]]+size[son[x][1]]+1;
}
inline int build()
{
    top=0;
    int n=strlen(s+1);
    for (int i=1;i<=n;i++)
    {
        key[++th]=s[i];
        val[th]=rand();
        stuck[top+1]=0;
        while (top>=1 && val[stuck[top]]<val[th])update(stuck[top--]);
        son[th][0]=stuck[top+1];
        if (top)son[stuck[top]][1]=th;
        stuck[++top]=th;
    }
    while (top)update(stuck[top--]);
    return stuck[1];
}

inline void copy(int x, int y)
{
    key[y]=key[x];
    son[y][0]=son[x][0];
    val[y]=val[x];
    son[y][1]=son[x][1];
    size[y]=size[x];
}
void breakup(int x,int a,int b,int k)
{
    if (k==0)
    {
        copy(x,b);
        return;
    }
    if (size[son[x][0]]>=k)
    {
        copy(x,b);
        breakup(son[x][0],a,son[b][0]=++th,k);
        update(b);
    }
    else
    {
        copy(x,a);
        breakup(son[x][1],son[a][1]=++th,b,k-size[son[x][0]]-1);
        update(a);
    }
}
void merge(int x, int a, int b)
{
    if (a==0)copy(b,x);
    else if (b==0)copy(a,x);
    else if (val[a]>val[b])
    {
        copy(a,x);
        merge(son[x][1]=++th,son[a][1],b);
        update(x);
    }
    else
    {
        copy(b,x);
        merge(son[x][0]=++th,a,son[b][0]);
        update(x);
    }

}
void out(int x)
{
    if (son[x][0])out(son[x][0]);
    if (key[x]=='c')dif++;
    printf("%c",key[x]);
    if (son[x][1])out(son[x][1]);
}

int main()
{
    srand(time(NULL));
    scanf("%d",&m);
    int pos,x,y,z,cnt=0,newtree;
    for (int i=1;i<=m;i++)
    {
        scanf("%d",&pos);
        if (pos<=2)cnt++,root[cnt]=root[cnt-1];
        if (pos==1)
        {
            scanf("%d%s",&x,s+1);
            x-=dif;
            newtree=build();
            int a=++th,b=++th,c=++th;
            breakup(root[cnt],a,b,x);
            merge(c,a,newtree);
            merge(root[cnt]=++th,c,b);
        }
        else if (pos==2)
        {
            scanf("%d%d", &x, &y);
            x -= dif, y -= dif;
            int a = ++th, b = ++th;
            breakup(root[cnt], a, b, x - 1);
            int c = ++th, d = ++th;
            breakup(b, c, d, y);
            merge(root[cnt] = ++th, a, d);
        }
        else
        {
            scanf("%d%d%d", &z, &x, &y);
            x -= dif, y -= dif, z -= dif;
            int a = ++th, b = ++th;
            breakup(root[z], a, b, x - 1);
            int c = ++th, d = ++th;
            breakup(b, c, d, y);
            out(c);
            printf("\n");
        }
    }

    return 0;
}
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ext/rope>

using namespace std;
using namespace __gnu_cxx;

const int maxn=50500;

crope tmp,var[maxn],t;
int n,op,p,c,v,d,curv;
char str[maxn];

int main()
{
  while(scanf("%d",&n)!=EOF)
  {
    curv=0;d=0;
    while(n--)
    {
      scanf("%d",&op);
      if(op==1)
      {
        scanf("%d%s",&p,str);
        p-=d;
        t.insert(p,str);
        var[++curv]=t;
      }
      else if(op==2)
      {
        scanf("%d%d",&p,&c);
        p-=d;c-=d;
        t.erase(p-1,c);
        var[++curv]=t;
      }
      else if(op==3)
      {
        scanf("%d%d%d",&v,&p,&c);
        v-=d;p-=d;c-=d;
        tmp=var[v].substr(p-1,c);
        d+=count(tmp.begin(),tmp.end(),'c');
        cout<<tmp<<endl;
      }
    }
  }
  return 0;
}