Lexicographical Substring Search (spoj7259) (sam(后缀自动机)+第k小子串)

Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string \(S\) and asked him \(Q\) questions of the form:

If all distinct substrings of string \(S\) were sorted lexicographically, which one will be the \(K-th\) smallest?

After knowing the huge number of questions Kinan will ask, Daniel figured out that he can't do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given \(S\) will answer Kinan's questions.

Example:

\(S\)= "aaa" (without quotes)

substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:

"a", "aa", "aaa".

Input

In the first line there is Kinan's string \(S\) (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer \(Q\) (\(Q<=500\)) , the number of questions Daniel will be asked. In the next \(Q\) lines a single integer \(K\) is given (\(0<K<2^{31}\)).

Output

Output consists of \(Q\) lines, the i-th contains a string which is the answer to the i-th asked question.

Example

Input:

aaa

2

2

3

Output:

aa

aaa

Edited: Some input file contains garbage at the end. Do not process them.

题意：

给定一个字符串，求排名第k小的串。

题解：

把串塞进一个后缀自动机，在图上反向拓扑求出每个点的后继串的数量，然后像在权值线段树上找第\(k\)大一样找就行了。

#include<bits/stdc++.h>

using namespace std;

const int N=2000010;

char s[N];

int a[N],c[N],ans[N];

struct SAM{

    int last,cnt;

    int size[N],ch[N][26],fa[N<<1],l[N<<1];

    void ins(int c){

        int p=last,np=++cnt;last=np;l[np]=l[p]+1;

        for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;

        if(!p)fa[np]=1;

        else{

            int q=ch[p][c];

            if(l[p]+1==l[q])fa[np]=q;

            else{

                int nq=++cnt;l[nq]=l[p]+1;

                memcpy(ch[nq],ch[q],sizeof ch[q]);

                fa[nq]=fa[q];fa[q]=fa[np]=nq;

                for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;

                size[nq]=1;

            }

        }

        size[np]=1;

    }

    void build(char s[]){

        int len=strlen(s+1);

        last=cnt=1;

        for(int i=1;i<=len;++i)ins(s[i]-'a');

    }

    void calc(int len){

        for(int i=1;i<=cnt;++i)c[l[i]]++;

        for(int i=1;i<=cnt;++i)c[i]+=c[i-1];

        for(int i=1;i<=cnt;++i)a[c[l[i]]--]=i;

        for(int i=cnt;i;--i){

            int p=a[i];

            for(int j=0;j<26;++j){

                size[p]+=size[ch[p][j]];

            }

        }

    }

    void find(int k){

        int p=1;

        while(k){

            int a=0;

            while(k>size[ch[p][a]]&&a<26){

                if (ch[p][a]) k-=size[ch[p][a]];

                a++;

            }

            putchar('a'+a);k--;

            p=ch[p][a];

        }

    }

}sam;

int main(){

    cin>>s+1;

    sam.build(s);

    sam.calc(strlen(s+1));

    int t;

    cin>>t;

    while(t--){

        int x;

        scanf("%d",&x);

        sam.find(x);putchar('\n');

    }

}

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