Description

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 stringSand asked himQquestions of the form:

If all distinct substrings of stringSwere sorted lexicographically, which one will be theK-thsmallest?

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 givenSwill 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".

Solution

在後綴自動機的DAG上DP，貪心地選取節點

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int n,las=1,tot=1,T,tot2,head[1800010],du[1800010],dp[1800010];
char s[900005];
struct SAM
{
    int ch[26],fa,len;
}sam[5000010];
struct Edge
{
    int to,nxt;
}edge[1800010];
queue
 
<int>q;
inline int read()
{
    int w=0,f=1;
    char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9')w=(w<<1)+(w<<3)+ch-'0',ch=getchar();
    return w*f;
}
void insert(int c)
{
    int p=las,np=las=++tot;
    sam[np].len=sam[p].len+1;
    for(;p&&!sam[p].ch[c];p=sam[p].fa) sam[p].ch[c]=np;
    if(!p) sam[np].fa=1;
    else
    {
        int q=sam[p].ch[c];
        if(sam[q].len==sam[p].len+1) sam[np].fa=q;
        else
        {
            int nq=++tot;
            sam[nq]=sam[q],sam[nq].len=sam[p].len+1,sam[q].fa=sam[np].fa=nq;
            for(;p&&sam[p].ch[c]==q;p=sam[p].fa) sam[p].ch[c]=nq;
        }
    }
}
int main()
{
    scanf("%s",s+1),n=strlen(s+1),T=read();
    for(int i=1;i<=n;i++) insert(s[i]-'a');
    for(int i=1;i<=tot;i++) for(int j=0;j<26;j++) if(sam[i].ch[j]) edge[++tot2]=(Edge){i,head[sam[i].ch[j]]},head[sam[i].ch[j]]=tot2,++du[i];
    for(int i=1;i<=tot;i++) if(!du[i]) q.push(i);
    while(q.size())
    {
        int u=q.front();
        q.pop(),dp[u]=1;
        for(int i=0;i<26;i++) if(sam[u].ch[i]) dp[u]+=dp[sam[u].ch[i]];
        for(int i=head[u];i;i=edge[i].nxt)
        {
            int v=edge[i].to;
            du[v]--;
            if(!du[v]) q.push(v);
        }
    }
    for(;T;T--)
    {
        int k=read(),p=1;
        while(k) for(int i=0;i<26;i++) if(sam[p].ch[i])
            if(dp[sam[p].ch[i]]>=k)
            {
                putchar(i+'a'),p=sam[p].ch[i],--k;break;
            }
            else k-=dp[sam[p].ch[i]];
        putchar(10);
    }
    return 0;
}