SPOJ 题目7528 SUBLEX - Lexicographical Substring Search（后缀自动机求排名k的子串）

SUBLEX - Lexicographical Substring Search

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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
这个oj我也是醉了，，刚开始还是用习惯的指针形式写后缀自动机，一直re，我以为太慢了，就换用数组的试了试，一直TLE，题目是输入一组，我就吧输入多组改成输入一组，提交wa了，没有超时，然后就想是不是之前的代码也行，去掉！=EOF那个提交就过了，然后改了改数组形式的，提交也过了
ac代码
指针形式



#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
#define N 500010
struct sam
{
sam *pre,*son[26];
int len,g;
}que
,*root,*tail,*b
;
int tot;
void add(int c,int l)
{
sam *p=tail,*np=&que[tot++];
np->len=l;
tail=np;
while(p&&p->son[c]==NULL)
{
p->son[c]=np;
p=p->pre;
}
if(p==NULL)
np->pre=root;
else
{
sam *q=p->son[c];
if(p->len+1==q->len)
np->pre=q;
else
{
sam *nq=&que[tot++];
*nq=*q;
nq->len=p->len+1;
np->pre=q->pre=nq;
while(p&&p->son[c]==q)
{
p->son[c]=nq;
p=p->pre;
}
}
}
}
char str
;
int cnt
,len;
void build()
{
int i,j;
tot=0;
root=tail=&que[tot++];
len=strlen(str);
for(i=0;i<len;i++)
add(str[i]-'a',i+1);
memset(cnt,0,sizeof(cnt));
for(i=0;i<tot;i++)
cnt[que[i].len]++;
for(i=1;i<=len;i++)
cnt[i]+=cnt[i-1];
for(i=0;i<tot;i++)
b[--cnt[que[i].len]]=&que[i];
for(i=tot-1;i>=0;i--)
{
sam *p=b[i];
if(b[i]==NULL)
continue;
p->g=1;
for(j=0;j<26;j++)
{
if(p->son[j])
p->g+=p->son[j]->g;
}
}
}
void query(int x)
{
sam *p=root;
int i,j;
while(x)
{
for(i=0;i<26;i++)
{
if(p->son[i]!=NULL)
{
if(p->son[i]->g>=x)
{
printf("%c",'a'+i);
p=p->son[i];
x--;
break;
}
else
x-=p->son[i]->g;
}
}
}
printf("\n");
}
int main()
{
scanf("%s",str);
build();
int q;
scanf("%d",&q);
while(q--)
{
int x;
scanf("%d",&x);
query(x);
}
}

数组形式


#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define N 200020
char str
;
int cot,last;
int fa
,son
[26],len
,b
,n;
int g
,cnt
;
void add(int c,int l)
{
int p,np;
p=last;
np=last=++cot;
len[np]=++n;
while(p&&!son[p][c])
{
son[p][c]=np;
p=fa[p];
}
if(!p)
{
fa[np]=1;
}
else
{
int q=son[p][c];
if(len[p]+1==len[q])
fa[np]=q;
else
{
int nq=++cot;
len[nq]=len[p]+1;
memcpy(son[nq],son[q],sizeof(son[q]));
fa[nq]=fa[q];
fa[q]=fa[np]=nq;
while(son[p][c]==q)
{
son[p][c]=nq;
p=fa[p];
}
}
}
}
void build()
{
cot=last=1;
n=0;
/*	memset(son,0,sizeof(son));
memset(len,0,sizeof(len));
memset(b,0,sizeof(b));
memset(fa,0,sizeof(fa));*/
int i,j;
//	int Len=strlen(str);
for(i=0;str[i];i++)
add(str[i]-'a',i+1);
for(i=1;i<=cot;i++)
{
cnt[len[i]]++;
}
for(i=1;i<=n;i++)
{
cnt[i]+=cnt[i-1];
}
for(i=1;i<=cot;i++)
{
b[cnt[len[i]]--]=i;
}
}
void init()
{
int i,j;
for(i=cot;i;i--)
{
g[b[i]]=1;
for(j=0;j<26;j++)
g[b[i]]+=g[son[b[i]][j]];
}
}
void query(int x)
{
int p=1;
while(x)
{
int i;
for(i=0;i<26;i++)
{
if(son[p][i])
{
if(g[son[p][i]]>=x)
{
printf("%c",'a'+i);
p=son[p][i];
x--;
break;
}
else
x-=g[son[p][i]];
}
}
}
printf("\n");
}
int main()
{
scanf("%s",str);
build();
init();
int q;
scanf("%d",&q);
while(q--)
{
int x;
scanf("%d",&x);
query(x);
}
}