spoj SUBLEX Lexicographical Substring Search【解法二】

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.

后缀自动机解法见【这里】。

建出来sa，把询问排序之后在sa上扫描一遍，每个后缀会贡献n−sa[i]+1−height[i]个本质不同的子串。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct query
{
int id,k;
bool operator < (const query &qq) const
{
return k<qq.k;
}
}a[510];
char s[200010];
int sa[200010],rank[200010],hei[200010],cnt[200010],f[200010],ansp[510],anslen[510],
n,m=26,q;
int main()
{
int p,now=0;
scanf("%s",s+1);
n=strlen(s+1);
for (int i=1;i<=n;i++) cnt[rank[i]=s[i]-'a'+1]++;
for (int i=2;i<=m;i++) cnt[i]+=cnt[i-1];
for (int i=n;i;i--) sa[cnt[rank[i]]--]=i;
for (int k=1;k<=n;k<<=1)
{
p=0;
for (int i=n-k+1;i<=n;i++) f[++p]=i;
for (int i=1;i<=n;i++) if (sa[i]>k) f[++p]=sa[i]-k;
for (int i=1;i<=m;i++) cnt[i]=0;
for (int i=1;i<=n;i++) cnt[rank[f[i]]]++;
for (int i=2;i<=m;i++) cnt[i]+=cnt[i-1];
for (int i=n;i;i--) sa[cnt[rank[f[i]]]--]=f[i];
for (int i=1;i<=n;i++) f[i]=rank[i];
rank[sa[1]]=1;
for (int i=2;i<=n;i++)
rank[sa[i]]=rank[sa[i-1]]+
(f[sa[i]]!=f[sa[i-1]]||f[sa[i]+k]!=f[sa[i-1]+k]);
if (rank[sa
]>=n) break;
m=rank[sa
];
}
for (int i=1;i<=n;i++)
{
hei[rank[i]]=hei[rank[i-1]];
if (hei[rank[i]]) hei[rank[i]]--;
while (s[i+hei[rank[i]]]==s[sa[rank[i]-1]+hei[rank[i]]]) hei[rank[i]]++;
}
scanf("%d",&q);
for (int i=1;i<=q;i++)
{
scanf("%d",&a[i].k);
a[i].id=i;
}
sort(a+1,a+q+1);
p=1;
for (int i=1;i<=q;i++)
{
while (now+n-sa[p]+1-hei[p]<a[i].k)
{
now+=n-sa[p]+1-hei[p];
p++;
}
ansp[a[i].id]=sa[p];
anslen[a[i].id]=hei[p]+a[i].k-now;
}
for (int i=1;i<=q;i++)
{
for (int j=ansp[i];j<=ansp[i]+anslen[i]-1;j++) putchar(s[j]);
putchar('\n');
}
}