URAL 1992 CVS

CVS

题目连接：

http://acm.timus.ru/problem.aspx?space=1&num=1992

Description

Yoda: Visit I will the cloners on Kamino... And see this army they have created for the Republic.
Cloners from the Kamino planet breed some of the finest clones. Such good results are due to the careful management over the clones’ evolution. The Kaminuans are now busy working out a new study technology that lets increase the clones’ effectiveness. The cloners have come up with a new control system CVS (Clone Version System) that makes managing the progress of experiments easier. The system is quite simple to use.
The Kaminuans have some set of educational programs at their disposal. A clone’s effectiveness depends on which programs and in which order he has learned. The Kaminuans can teach any clone a program as long as this clone hasn’t learned it already.
To make the experiments even easier to conduct, the Kaminuans enabled canceling the changes made by the last program the clone has learned. In this case, the clone’s knowledge returns to the level when the program hasn’t yet been studied. Then this clone can study this program in the future. You can cancel programs at any time unless the clone is at the basic knowledge level.
Besides the ‘roll back’ function, a clone can ‘re-learn’ a program. If one cancels some program by mistake, he can cancel the cancellation. The CVS keeps record of each clone’s canceled programs. After a program is canceled, the CVS adds another record. After a clone re-learn a program, the record is deleted. If a clone learn (not relearn) a program, all cancellation record history of this clone is deleted. You can use the re-learn function as long as the record history for this clone contains any records.
Finally, the system has a ‘clone’ option. If a Kaminuan likes the current variant of a clone, he can clone the clone, that is, create a new clone with the same sequence of taught programs and cancellation history.
Initially the Kaminuans have a single clone with basic knowledge. Help them analyze the progress of the experiments.

Input

The first line of the input contains numbers n — the number of queries — and m — the number of educational programs (1 ≤ n, m ≤ 5·10 5). Each of the following n lines has one of the formats given below.
learn ci pi. Teach clone ci program pi (1 ≤ pi ≤ m).
rollback ci. Cancel the last learned program for clone ci.
relearn ci. Apply ‘re-learn’ function to clone ci.
clone ci. Clone the clone ci.
check ci. Display the last program clone ci has learned and knows at the moment.
It is guaranteed that rollback won’t be applied to the clone that is at the basic knowledge level. learn is always applied with the program a clone doesn’t already know. relearn is only applied if the cancellation history of a clone is not empty. It is also guaranteed that only the clones that already exist can occur in the queries. The numbers are assigned to the clones in the order the clones appear. The Kaminuans started their experiments from clone number one.

Output

For each check ci query display the result on a single line. If some clone has only basic knowledge, print basic, otherwise print the number of the last learned program.

Sample Input

9 10
learn 1 5
learn 1 7
rollback 1
check 1
clone 1
relearn 2
check 2
rollback 1
check 1

Sample Output

5
7
basic

Hint

题意

有五个操作

learn x y，让x学习y知识

rollback x，让x遗忘最后一个学习的知识

check x，问x最后学习的是啥

relearn x，让x学习一个最后一个遗忘的知识

clone x，复制一个

题解：

实际上就是维护两个可持久化的栈，因为有克隆和回到之前的操作嘛

我们去维护两棵树，一个是学习的树，一个是遗忘的树

然后根据标记去维护就好了

这道题比较毒瘤，必须加读入挂才能过……

代码

#include<bits/stdc++.h>
using namespace std;
namespace fastIO{
    #define BUF_SIZE 100000
    #define OUT_SIZE 100000
    #define ll long long
    //fread->read
    bool IOerror=0;
    inline char nc(){
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if (p1==pend){
            p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if (pend==p1){IOerror=1;return -1;}
            //{printf("IO error!\n");system("pause");for (;;);exit(0);}
        }
        return *p1++;
    }
    inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
    inline void read(int &x){
        bool sign=0; char ch=nc(); x=0;
        for (;blank(ch);ch=nc());
        if (IOerror)return;
        if (ch=='-')sign=1,ch=nc();
        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
        if (sign)x=-x;
    }
    inline void read(ll &x){
        bool sign=0; char ch=nc(); x=0;
        for (;blank(ch);ch=nc());
        if (IOerror)return;
        if (ch=='-')sign=1,ch=nc();
        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
        if (sign)x=-x;
    }
    inline void read(double &x){
        bool sign=0; char ch=nc(); x=0;
        for (;blank(ch);ch=nc());
        if (IOerror)return;
        if (ch=='-')sign=1,ch=nc();
        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
        if (ch=='.'){
            double tmp=1; ch=nc();
            for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
        }
        if (sign)x=-x;
    }
    inline void read(char *s){
        char ch=nc();
        for (;blank(ch);ch=nc());
        if (IOerror)return;
        for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
        *s=0;
    }
    inline void read(char &c){
        for (c=nc();blank(c);c=nc());
        if (IOerror){c=-1;return;}
    }
    //getchar->read
    inline void read1(int &x){
        char ch;int bo=0;x=0;
        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
        if (bo)x=-x;
    }
    inline void read1(ll &x){
        char ch;int bo=0;x=0;
        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
        if (bo)x=-x;
    }
    inline void read1(double &x){
        char ch;int bo=0;x=0;
        for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
        for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
        if (ch=='.'){
            double tmp=1;
            for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
        }
        if (bo)x=-x;
    }
    inline void read1(char *s){
        char ch=getchar();
        for (;blank(ch);ch=getchar());
        for (;!blank(ch);ch=getchar())*s++=ch;
        *s=0;
    }
    inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
    //scanf->read
    inline void read2(int &x){scanf("%d",&x);}
    inline void read2(ll &x){
        #ifdef _WIN32
            scanf("%I64d",&x);
        #else
        #ifdef __linux
            scanf("%lld",&x);
        #else
            puts("error:can't recognize the system!");
        #endif
        #endif
    }
    inline void read2(double &x){scanf("%lf",&x);}
    inline void read2(char *s){scanf("%s",s);}
    inline void read2(char &c){scanf(" %c",&c);}
    inline void readln2(char *s){gets(s);}
    //fwrite->write
    struct Ostream_fwrite{
        char *buf,*p1,*pend;
        Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}
        void out(char ch){
            if (p1==pend){
                fwrite(buf,1,BUF_SIZE,stdout);p1=buf;
            }
            *p1++=ch;
        }
        void print(int x){
            static char s[15],*s1;s1=s;
            if (!x)*s1++='0';if (x<0)out('-'),x=-x;
            while(x)*s1++=x%10+'0',x/=10;
            while(s1--!=s)out(*s1);
        }
        void println(int x){
            static char s[15],*s1;s1=s;
            if (!x)*s1++='0';if (x<0)out('-'),x=-x;
            while(x)*s1++=x%10+'0',x/=10;
            while(s1--!=s)out(*s1); out('\n');
        }
        void print(ll x){
            static char s[25],*s1;s1=s;
            if (!x)*s1++='0';if (x<0)out('-'),x=-x;
            while(x)*s1++=x%10+'0',x/=10;
            while(s1--!=s)out(*s1);
        }
        void println(ll x){
            static char s[25],*s1;s1=s;
            if (!x)*s1++='0';if (x<0)out('-'),x=-x;
            while(x)*s1++=x%10+'0',x/=10;
            while(s1--!=s)out(*s1); out('\n');
        }
        void print(double x,int y){
            static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,
                1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,
                100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};
            if (x<-1e-12)out('-'),x=-x;x*=mul[y];
            ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;
            ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);
            if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}
        }
        void println(double x,int y){print(x,y);out('\n');}
        void print(char *s){while (*s)out(*s++);}
        void println(char *s){while (*s)out(*s++);out('\n');}
        void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}
        ~Ostream_fwrite(){flush();}
    }Ostream;
    inline void print(int x){Ostream.print(x);}
    inline void println(int x){Ostream.println(x);}
    inline void print(char x){Ostream.out(x);}
    inline void println(char x){Ostream.out(x);Ostream.out('\n');}
    inline void print(ll x){Ostream.print(x);}
    inline void println(ll x){Ostream.println(x);}
    inline void print(double x,int y){Ostream.print(x,y);}
    inline void println(double x,int y){Ostream.println(x,y);}
    inline void print(char *s){Ostream.print(s);}
    inline void println(char *s){Ostream.println(s);}
    inline void println(){Ostream.out('\n');}
    inline void flush(){Ostream.flush();}
    //puts->write
    char Out[OUT_SIZE],*o=Out;
    inline void print1(int x){
        static char buf[15];
        char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
        while(x)*p1++=x%10+'0',x/=10;
        while(p1--!=buf)*o++=*p1;
    }
    inline void println1(int x){print1(x);*o++='\n';}
    inline void print1(ll x){
        static char buf[25];
        char *p1=buf;if (!x)*p1++='0';if (x<0)*o++='-',x=-x;
        while(x)*p1++=x%10+'0',x/=10;
        while(p1--!=buf)*o++=*p1;
    }
    inline void println1(ll x){print1(x);*o++='\n';}
    inline void print1(char c){*o++=c;}
    inline void println1(char c){*o++=c;*o++='\n';}
    inline void print1(char *s){while (*s)*o++=*s++;}
    inline void println1(char *s){print1(s);*o++='\n';}
    inline void println1(){*o++='\n';}
    inline void flush1(){if (o!=Out){if (*(o-1)=='\n')*--o=0;puts(Out);}}
    struct puts_write{
        ~puts_write(){flush1();}
    }_puts;
    inline void print2(int x){printf("%d",x);}
    inline void println2(int x){printf("%d\n",x);}
    inline void print2(char x){printf("%c",x);}
    inline void println2(char x){printf("%c\n",x);}
    inline void print2(ll x){
        #ifdef _WIN32
            printf("%I64d",x);
        #else
        #ifdef __linux
            printf("%lld",x);
        #else
            puts("error:can't recognize the system!");
        #endif
        #endif
    }
    inline void println2(ll x){print2(x);printf("\n");}
    inline void println2(){printf("\n");}
    #undef ll
    #undef OUT_SIZE
    #undef BUF_SIZE
};
using namespace fastIO;
const int maxn = 5e5+7;
int pos1[maxn],pos2[maxn];
int fa[maxn],fa2[maxn];
vector<int>G1[maxn],G2[maxn];
int T1[maxn],T2[maxn];
int tot1=0,tot2=0,num=2;
void learn()
{
    int x,y;
    scanf("%d%d",&x,&y);
    T1[++tot1]=y;
    G1[pos1[x]].push_back(tot1);
    fa[tot1]=pos1[x];
    pos1[x]=tot1;
}
void rollback()
{
    int x;
    scanf("%d",&x);
    T2[++tot2]=pos1[x];
    G2[pos2[x]].push_back(tot2);
    fa2[tot2]=pos2[x];
    pos2[x]=tot2;
    pos1[x]=fa[pos1[x]];
}
void check()
{
    int x;
    scanf("%d",&x);
    if(pos1[x]==0)printf("basic\n");
    else printf("%d\n",T1[pos1[x]]);
}
void clone()
{
    int x;
    scanf("%d",&x);
    pos1[num]=pos1[x],pos2[num]=pos2[x];
    num++;
}
void relearn()
{
    int x;
    scanf("%d",&x);
    pos1[x]=T2[pos2[x]];
    pos2[x]=fa2[pos2[x]];
}
char s[15];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        read1(s);
        if(s[0]=='l')learn();
        if(s[0]=='r'&&s[1]=='o')rollback();
        if(s[0]=='c'&&s[1]=='h')check();
        if(s[0]=='c'&&s[1]=='l')clone();
        if(s[0]=='r'&&s[1]=='e')relearn();
    }
}